Codeforces Round #440 (Div. 2) A,B,C

A. Search for Pretty Integers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two lists of non-zero digits.

Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.

The second line contains n distinct digits a1, a2, ..., an (1 ≤ ai ≤ 9) — the elements of the first list.

The third line contains m distinct digits b1, b2, ..., bm (1 ≤ bi ≤ 9) — the elements of the second list.

Output

Print the smallest pretty integer.

Examples

input

2 3
4 2
5 7 6

output

25

input

8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1

output

1

Note

In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.

In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.

 水题

#include<bits/stdc++.h>
using namespace std;
const int MAX = 20;
int main()
{
    int n,m,a[MAX],b[MAX],ans=999,i,j;
    cin>>n>>m;
    for(i=1;i<=n;i++) cin>>a[i];
    for(i=1;i<=m;i++) cin>>b[i];
    sort(a+1,a+1+n);
    sort(b+1,b+1+m);
    for(i=1;i<=n;i++)
    for(j=1;j<=m;j++)
    {
        if(a[i]==b[j])ans=min(ans,a[i]);
    }
    if(ans==999)
    {
        int u=max(a[1],b[1]),v=min(a[1],b[1]);
        ans=u+v*10;
    }
    cout<<ans<<endl;
}

B. Maximum of Maximums of Minimums

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array a1, a2, ..., an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?

Definitions of subsegment and array splitting are given in notes.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤  105) — the size of the array a and the number of subsegments you have to split the array to.

The second line contains n integers a1,  a2,  ...,  an ( - 109  ≤  ai ≤  109).

Output

Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.

Examples

input

5 2
1 2 3 4 5

output

5

input

5 1
-4 -5 -3 -2 -1

output

-5

Note

A subsegment [l,  r] (l ≤ r) of array a is the sequence al,  al + 1,  ...,  ar.

Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], ..., [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is k sequences (al1, ..., ar1), ..., (alk, ..., ark).

In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.

In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is min( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.

 水题.

1. 当n=1的时候绝对是取这一串序列的最小值.

2. 当n=2是会分成两个区间，但不管怎么分是最左边与最右边的数一定是不同区间的，可以推一下，如果要取每个区间最小值中最大的，如序列3 4 1 3 2   / 代表分区间，会有四种分法：3 / 4 1 3 2    3 4 / 1 3 2   3 4 1 / 3 2    3 4 1 3 / 2  ，如果a[1]>a[0],那么最小的还是a[0]， 没有变化，如果a[1]<a[0],区间最小值变小,无意义,，从后往前推也是这个道理，所以当n=2的情况下，只需要比较最左边与最右边就可以了.

3.当n=3时，可以直接取到最大的数。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long m,n,maxx=-10000000090,minn=10000000900,i,x,a[100009];
    cin>>m>>n;
    if(n==1){
        for(i=0;i<m;i++){
            cin>>x;
            if(x<minn)
                minn=x;
        }
        cout<<minn<<endl;
    }
    else if(n==2){
        for(i=0;i<m;i++){
            cin>>a[i];
        }
        cout<<max(a[0],a[m-1])<<endl;
    }
    else {
        for(i=0;i<m;i++){
            cin>>x;
            if(x>maxx)
                maxx = x;
        }
        cout<<maxx<<endl;
    }
    return 0;
}

C. Maximum splitting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input

The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.

Output

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples

input

1
12

output

3

input

2
6
8

output

1
2

input

3
1
2
3

output

-1
-1
-1

Note

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can't be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

 题目很简单，一开始没看懂题意，题意是求一个数n最多能分解成多少个非素数

优先考虑4，让其对4取模，这样就只有四种情况

余数为0时，正好整除必为多个4组成，这样就是最优解

余数为1时，4+4+1 = 9，退一位，n/4-1,余数1与两个 4凑成9，非素数

余数为2时，4+2=6，不变，4与余数2凑成6，非素数

余数为3时，4+4+4+3=6+9，退一位，（满足你>=15）

综上只要满足>=15,公式就可以通用

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,m,ans,flag;
    cin>>n;
    while(n--)
    {
        scanf("%d",&m);
        if(m==11)printf("-1
");
        else
        {
            flag=m%4;
            if(flag==0)ans=m/4;
            else if(flag==1)ans=m/4-1;
            else if(flag==2)ans=m/4;
            else ans=m/4-1;
            if(ans<=0)
            ans=-1;
            printf("%d
",ans);
        }
    }
}