448D

D. Multiplication Table

time limit per test

1 second

memory limit per test

256 megabytes

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.

Sample test(s)

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

/*
ID: LinKArftc
PROG: d.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <climits>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

ll n, m, k;

ll check(ll x) {
    ll ret = 0;
    for (int i = 1; i <= n; ++ i) {
        ll tmp = min(i * m, x);
        ret += tmp / i;
    }
    return ret;
}

int main() {

    while (~scanf("%I64d %I64d %I64d", &n, &m, &k)) {
        ll l = 1, r = n * m;
        while (l < r) {
            ll mid = (l + r) >> 1;
            ll sum = check(mid);
            if (sum >= k) r = mid;
            else l = mid + 1;
        }
        printf("%I64d
", r);
    }

    return 0;
}