Good Bye 2017

太菜了啊，一不小心就goodbye rating了

A. New Year and Counting Cards

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Your friend has n cards.

You know that each card has a lowercase English letter on one side and a digit on the other.

Currently, your friend has laid out the cards on a table so only one side of each card is visible.

You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.

For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.

To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.

Input

The first and only line of input will contain a string s (1 ≤ |s| ≤ 50), denoting the sides of the cards that you can see on the table currently. Each character of s is either a lowercase English letter or a digit.

Output

Print a single integer, the minimum number of cards you must turn over to verify your claim.

Examples

input

ee

output

2

input

z

output

0

input

0ay1

output

2

Note

In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.

In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.

In the third sample, we need to flip the second and fourth cards.

原谅我英语不好，看懂了就是统计元音和奇数个数

#include<bits/stdc++.h>
using namespace std;
string c="aeiou13579";
int main()
{
    string s;
    cin>>s;
    int ans=0;
    for(int i=0;s[i];i++)
        for(int j=0;c[j];j++)
            if(s[i]==c[j])
                ans++;
    cout<<ans;
    return 0;
}

B. New Year and Buggy Bot

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob programmed a robot to navigate through a 2d maze.

The maze has some obstacles. Empty cells are denoted by the character '.', where obstacles are denoted by '#'.

There is a single robot in the maze. It's start position is denoted with the character 'S'. This position has no obstacle in it. There is also a single exit in the maze. It's position is denoted with the character 'E'. This position has no obstacle in it.

The robot can only move up, left, right, or down.

When Bob programmed the robot, he wrote down a string of digits consisting of the digits 0 to 3, inclusive. He intended for each digit to correspond to a distinct direction, and the robot would follow the directions in order to reach the exit. Unfortunately, he forgot to actually assign the directions to digits.

The robot will choose some random mapping of digits to distinct directions. The robot will map distinct digits to distinct directions. The robot will then follow the instructions according to the given string in order and chosen mapping. If an instruction would lead the robot to go off the edge of the maze or hit an obstacle, the robot will crash and break down. If the robot reaches the exit at any point, then the robot will stop following any further instructions.

Bob is having trouble debugging his robot, so he would like to determine the number of mappings of digits to directions that would lead the robot to the exit.

Input

The first line of input will contain two integers n and m (2 ≤ n, m ≤ 50), denoting the dimensions of the maze.

The next n lines will contain exactly m characters each, denoting the maze.

Each character of the maze will be '.', '#', 'S', or 'E'.

There will be exactly one 'S' and exactly one 'E' in the maze.

The last line will contain a single string s (1 ≤ |s| ≤ 100) — the instructions given to the robot. Each character of s is a digit from 0 to 3.

Output

Print a single integer, the number of mappings of digits to directions that will lead the robot to the exit.

Examples

input

5 6
.....#
S....#
.#....
.#....
...E..
333300012

output

1

input

6 6
......
......
..SE..
......
......
......
01232123212302123021

output

14

input

5 3
...
.S.
###
.E.
...
3

output

0

Note

For the first sample, the only valid mapping is , where D is down, L is left, U is up, R is right.

 

 全排列24种情况啊，等等为毛我写的只有23种，挂了两个题果断GG（do，while都不懂的我果断GG啊，但是我感觉思路很简洁啊，就是让1 2 3 4分别去代表U D L R，然后我a[0]代表上，a[1]代表下，a[2]代表左，a[3]代表右

#include<bits/stdc++.h>
using namespace std;
string s[51];
int a[6],sx,sy;
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        cin>>s[i];
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        if(s[i][j]=='S')sx=i,sy=j;
    string c;
    cin>>c;
    for(int i=0;i<4;i++)
        a[i]=i+48;
    int ans=0;
    do
    {
        int f=0;
        int tx=sx,ty=sy;
        for(int i=0;c[i];i++)
        {
            if(c[i]==a[0])
                tx-=1;
            else if(c[i]==a[1])
                tx+=1;
            else if(c[i]==a[2])
                ty-=1;
            else ty+=1;
            if(tx<0||ty<0||tx>=n||ty>=m)
                break;
            if(s[tx][ty]=='#')break;
            if(s[tx][ty]=='E'){f=1;break;}
        }
        ans+=f;
    }while(next_permutation(a,a+4));
    cout<<ans;
    return 0;
}

C. New Year and Curling

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Carol is currently curling.

She has n disks each with radius r on the 2D plane.

Initially she has all these disks above the line y = 10100.

She then will slide the disks towards the line y = 0 one by one in order from 1 to n.

When she slides the i-th disk, she will place its center at the point (xi, 10100). She will then push it so the disk’s ycoordinate continuously decreases, and x coordinate stays constant. The disk stops once it touches the line y = 0or it touches any previous disk. Note that once a disk stops moving, it will not move again, even if hit by another disk.

Compute the y-coordinates of centers of all the disks after all disks have been pushed.

Input

The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.

The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks.

Output

Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.

Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker program will consider your answer correct if  for all coordinates.

Example

input

6 2
5 5 6 8 3 12

output

2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613

Note

The final positions of the disks will look as follows:

In particular, note the position of the last disk.

C我逃了开根为负的情况，但是也不会是0啊，最起码是r

#include<bits/stdc++.h>
using namespace std;
int x[1005];
double y[1008];
int main()
{
    int n,r;
    cin>>n>>r;
    for(int i=0;i<n;i++)
        cin>>x[i];
    for(int i=0;i<n;i++)
    {
        y[i]=r;
        for(int j=0;j<i;j++)
           if(fabs(x[j]-x[i])<=2.*r)
           y[i]=max(y[i],y[j]+sqrt(4*r*r-(x[i]-x[j])*(x[i]-x[j])));
    }
    for(int i=0;i<n;i++)
    printf("%.12f ",y[i]);
    return 0;
}

D. New Year and Arbitrary Arrangement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given three integers k, pa and pb.

You will construct a sequence with the following algorithm: Initially, start with the empty sequence. Each second, you do the following. With probability pa / (pa + pb), add 'a' to the end of the sequence. Otherwise (with probability pb / (pa + pb)), add 'b' to the end of the sequence.

You stop once there are at least k subsequences that form 'ab'. Determine the expected number of times 'ab' is a subsequence in the resulting sequence. It can be shown that this can be represented by P / Q, where P and Q are coprime integers, and . Print the value of .

Input

The first line will contain three integers integer k, pa, pb (1 ≤ k ≤ 1 000, 1 ≤ pa, pb ≤ 1 000 000).

Output

Print a single integer, the answer to the problem.

Examples

input

1 1 1

output

2

input

3 1 4

output

370000006

Note

The first sample, we will keep appending to our sequence until we get the subsequence 'ab' at least once. For instance, we get the sequence 'ab' with probability 1/4, 'bbab' with probability 1/16, and 'aab' with probability 1/8. Note, it's impossible for us to end with a sequence like 'aabab', since we would have stopped our algorithm once we had the prefix 'aab'.

The expected amount of times that 'ab' will occur across all valid sequences is 2.

For the second sample, the answer is equal to .

#include<bits/stdc++.h>
using namespace std;
const int MD=1e9+7,N=2005;
int dp[N][N];
int po(int a,int x)
{
    int ans=1;
    while(x)
    {
        if(x&1) ans=1LL*ans*a%MD;
        a=1LL*a*a%MD;
        x>>=1;
    }
    return ans;
}
int main()
{
    int k,a,b,ta,tb,ans=0;
    scanf("%d%d%d",&k,&a,&b);
    ta=1LL*a*po(a+b,MD-2)%MD;
    tb=MD-ta+1;
    dp[1][0]=1;
    for(int i=1; i<=k; i++)
        for(int j=0; j<k-i; j++)
        {
            dp[i+1][j]=(dp[i+1][j]+1LL*dp[i][j]*ta%MD)%MD;
            if(j+i+i>=k)ans=(ans+1LL*dp[i][j]*tb%MD*(j+i+i-k)%MD)%MD;
            else dp[i][j+i]=(dp[i][j+i]+1LL*dp[i][j]*tb%MD)%MD;
        }
    ans=(ans+k+1LL*a*po(b,MD-2)%MD)%MD;
    printf("%d
",ans);
}