//本题思想：二分
/*
设第k个数为n，另设在n之前的整数个数为res
对于n的左半部的数：res < k  对于n的右半部的数：res >= k
由此二分
*/

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 5e5 + 10;

LL n,m;  //若开int，需改为: (1) min((LL)m,mid/i)  (2) r = (LL)n*m;
LL k;

bool check(LL mid){

    LL res = 0;
    for(int i=1;i<=n;i++){   //n行m列
        res += min(m,mid/i);  //逐行计算满足二分条件的个数：要么整行都满足，要么取[mid/i]
    }

    return res >= k;
}

int main(){

    cin >> n >> m >> k;

    LL l = 1, r = n*m;
    while(l<r){
        LL mid = l + r >> 1;
        if(check(mid)){
            r = mid;
        }else{
            l = mid + 1;
        }
    }

    printf("%lld",r);


    return 0;
}