time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ann × 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 32 4 6

如果直接进行搜索，时间复杂度是O(n*m)，肯定会超时。利用在乘法表中，每一行的数字都是(1*m,2*m,3*m,……,n*m)(m是行号)这一特点对行进行二分，mid/m就是mid

在这行中的排名。通过对每行进行搜索，就可以知道mid在鼠标中的总排名。

#include 
    
     #include 
     
      #include 
      
       using namespace std;long long t, n, m;bool judge(long long x) {    long long cnt = 0;    for (long long i = 1; i <= m; i++) {    	//注意x/i 大于n的情况         cnt += min(x/i, n);    }    return cnt >= t;}int main() {    while (scanf("%lld%lld%lld", &n, &m, &t) != EOF) {        long long lb = 1, ub = m*n;        long long ans = 0;        while (ub - lb >= 0) {            long long mid = (lb + ub)>>1;            if (judge(mid)) {                ans = mid;                ub = mid - 1;            }            else {                lb = mid + 1;            }        }        printf("%lld\n", ans);    }    return 0;}