Codeforces Round #395 (Div. 2) A. Taymyr is calling you【数论/最小公倍数】

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 everym 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.

【题意】：某人每间隔n分钟打一次房间内电话，而一些人每间隔m分钟就进入房间，要阻挠多少个人才能保证在一定时间段z内电话无人接听？

---

让你求1-z之间有多少数既是n的倍数又是m的倍数（n，m的最小公倍数）。

【代码】：

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f

int main()
{
    int n,m,k;
    while(cin>>n>>m>>k)
    {
        cout<<k/(n/(__gcd(n,m))*m)<<endl;
    }
    return 0;
}