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  
      
    
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Os")
using namespace std;
template <typename T, typename K>
inline void umax(T &a, K b) {
  a = max(a, (T)b);
}
template <typename T, typename K>
inline void umin(T &a, K b) {
  a = min(a, (T)b);
}
const int32_t N = 8e6;
const long long INF = 1e16 + 10;
const long double EPS = 1e-6;
const long long II = 1e9 + 10;
const long long MOD = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  long long ll = 0, rr = n * m + 1;
  while (ll < rr) {
    long long mm = (ll + rr) >> 1LL;
    long long cnt = 0;
    for (long long i = 1; i <= n; i++) {
      cnt += min(m, mm / i);
    }
    if (cnt < k)
      ll = mm + 1;
    else
      rr = mm;
  }
  cout << ll << endl;
}
