CodeForces 448D Multiplication Table

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.

Example

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

能推出来一个规律：n是行，m是列，对于一个数x，它在第i行中，有(x/i)个数字小于等于x
确定第几个数之后，我们枚举x就可以了。具体请看代码：

#include <stdio.h>
#define ll long long
ll m, n, k;
ll solve(ll a) {
	ll sum = 0;
	for(int i = 1; i <= n; i++) {
		if(a/i > m) sum += m;
		else sum += a/i; 
	}
	return sum;
}
int main() {
	ll r, l, mid;
	while(~scanf("%lld %lld %lld", &n, &m, &k)) {
		l = 0; r = k;
		while(l < r) {
			mid = l - (l - r)/2;
			if(solve(mid) >= k) r = mid;
			else l = mid+1;
		}
		printf("%lld\n", l);
	}
	return 0;
}