題干：

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

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.

題目大意：

? ? 對(duì)于給定的一個(gè)序列進(jìn)行操作，序列長(zhǎng)度為n，進(jìn)行操作，第i步操作即是把從第i起到第n-i+1的數(shù)進(jìn)行翻轉(zhuǎn)，保證2*i<=n+1;并且進(jìn)行輸出

解題報(bào)告：

? ? 直接看樣例，可以找規(guī)律。。。（但是為什么可以這樣？）

AC代碼：

#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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