【CodeForces - 764A】Taymyr is calling you （找规律，水题）

题干：

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

Input

1 2 5

Output

Input

2 3 9

Output

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.

题目大意：

对于给定的一个序列进行操作，序列长度为n，进行操作，第i步操作即是把从第i起到第n-i+1的数进行翻转，保证2*i<=n+1;并且进行输出

解题报告：

直接看样例，可以找规律。。。（但是为什么可以这样？）

AC代码：

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

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【CodeForces - 764A】Taymyr is calling you （找规律，水题）

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