Multiplication Table 【二分】

Multiplication Table

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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.

Sample Input

Input

2 2 2

Output

2

Input

2 3 4

Output

3

Input

1 10 5

Output

5

Hint

A 2 × 3 multiplication table looks like this:

1 2 3
2 4 6
思路：对乘法表中元素进行二分，二分的条件是对每一个mid求出每一行比它小的个数和然后与k比较；
代码：
#include<stdio.h>#include<math.h>#include<algorithm>#include<string.h>using namespace std;long long n,m,k;bool judge(long long mid){long long sum=0;for(long long i=1;i<=n;i++){sum=sum+min(m,mid/i);}return sum>=k;}int main(){while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF){long long ans=0;long long l=1,r=n*m;while(l<=r){long long mid=(l+r)/2;if(judge(mid)){ans=mid;r=mid-1;}else{l=mid+1;}}printf("%lld\n",ans);}return 0; }