CF448D：Multiplication Table（二分）

D. Multiplication Table

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 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

题意：大概意思是一个n*m的二位数组，dp[i][j] = i*j，将数组所有的数升序排序后，求第k个数。

思路：从1~n*m二分，遍历每一行求有多少个数小于等于mid的跟k判断一下就行。

# include <stdio.h>
# include <algorithm>
using namespace std;
typedef long long LL;
LL n, m, k;
bool fun(LL num)
{
    LL ans = 0;
    for(LL i=1; i<=n; ++i)
    {
        LL imin = min(i*m, num);
        ans += imin/i;
    }
    return ans < k;
}
int main()
{
    while(~scanf("%I64d%I64d%I64d",&n,&m,&k))
    {
        LL l = 0, r = n*m, mid;
        while(l<r)
        {
            mid = (l+r)>>1;
            if(fun(mid))
                l = mid+1;
            else
                r = mid;
        }
        printf("%I64d\n",r);
    }
    return 0;
}