Codeforces Round #395 (Div. 2)(A.思维，B,水)

A. Taymyr is calling you

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.

Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

Input

The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 10⁴).

Output

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

Examples

Input

1 1 10

Output

10

Input

1 2 5

Output

2

Input

2 3 9

Output

1

Note

Taymyr is a place in the north of Russia.

In the first test the artists come each minute, as well as the calls, so we need to kill all of them.

In the second test we need to kill artists which come on the second and the fourth minutes.

In the third test — only the artist which comes on the sixth minute.

题目链接：http://codeforces.com/contest/764/problem/A

分析：此题竟然是求n与m的最小公倍数，我TM是智障了！试了三次，开始以为就是ｔ／（m×n），智障宝宝！

不说太多，都是泪啊！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int n,m,t;
    while(cin>>n>>m>>t)
    {
        int x=gcd(n,m);
        int y=n*m/x;
        int z=t/y;
        cout<<z<<endl;
    }
    return 0;
}

B. Timofey and cubes

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number a_i, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.

After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

Input

The first line contains single integer n (1 ≤ n ≤ 2·10⁵) — the number of cubes.

The second line contains n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹), where a_i is the number written on the i-th cube after Dima has changed their order.

Output

Print n integers, separated by spaces — the numbers written on the cubes in their initial order.

It can be shown that the answer is unique.

Examples

Input

7 
4 3 7 6 9 1 2

Output

2 3 9 6 7 1 4

Input

8 
6 1 4 2 5 6 9 2

Output

2 1 6 2 5 4 9 6

Note

Consider the first sample.

1. At the begining row was [2, 3, 9, 6, 7, 1, 4].
2. After first operation row was [4, 1, 7, 6, 9, 3, 2].
3. After second operation row was [4, 3, 9, 6, 7, 1, 2].
4. After third operation row was [4, 3, 7, 6, 9, 1, 2].
5. At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

               

       

题目链接：http://codeforces.com/contest/764/problem/B

分析：智障宝宝继续犯傻，看了半天，以为规律就是奇偶变换，结果样例都没过，再回头看了下题，傻了眼，原来就是每两项第i项和第n－i＋１项交换！

智障宝宝第二摔！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
int a[200005];
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
            cin>>a[i];
            for(int i=1;i<=(n+1)/2;i+=2)
            {
                swap(a[i],a[n-i+1]);
            }
        for(int i=1;i<=n-1;i++)
            cout<<a[i]<<" ";
        cout<<a[n]<<endl;
    }
    return 0;
}