Problem Description

Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.

Input

The single line contains integers n, m and k (1 ≤ n, m ≤ 5·10⁵; 1 ≤ k ≤ n·m).

Output

Print the k-th largest number in a n × m multiplication table.

Examples

Input

2 2 2

Output

2

Input

2 3 4

Output

3

Input

1 10 5

Output

5

Note

A 2 × 3 multiplication table looks like this:

1 2 3
2 4 6
使用二分在1~n*m这个区间上寻找第k大的数。
即计算有多少数小于等于这个数并记录为cnt。找到最小的比第k大的cnt，该cnt所最对应的数即为答案。
计算cnt的方法为逐行cnt+=min(m,mid/i);(m为列数，mid为值，i为行数)
AC代码如下：

#include<cstdio>
#include<algorithm>
using namespace std;
long long n,m,l,r,mid,k,cnt,templ,tempr,tempm;
int main(){
    scanf("%lld%lld%lld",&n,&m,&k);
    l=1,r=n*m;
    while(r>=l){
        mid=(l+r)/2;
        cnt=0;
        for(long long i=1;i<=n;i++){
            cnt+=min(m,mid/i);
        }
        if(cnt>=k)
            r=mid-1;
        else
            l=mid+1;
    }
    printf("%lld",l);
}