Day8 - D - Multiplication Table CodeForces - 448D

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
 
思路：二分套二分，二分每个数，求出是第几大，每次比较一列，该列中比他小的数就是min(x,i*n)/i(i表示第几列)

typedef long long LL;
typedef pair<LL, LL> PLL;
 
LL n, m, k;
 
LL check(LL x) {
    LL ret = , b;
    for(LL i = ; i <= m; ++i) {
        b = min(x, i*n);
        ret += b/i;
    }
    return ret;
}
 
int main() {
    ios::sync_with_stdio(false), cin.tie();
    cin >> n >> m >> k;
    LL l = , r = n*m, mid, ans;
    while(l <= r) {
        mid = (l + r) >> ;
        if(check(mid) >= k) {
            ans = mid;
            r = mid - ;
        } else
            l = mid + ;
    }
    cout << ans << "\n";
    return ;
}