codeforces 762A k-th divisor

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题目描述：

You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2

Output

2

Input

5 3

Output

-1

Input

12 5

Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

题目大意：

找<=n的第k个因子

解题思路：

扫描到sqrt（n),i是n的因子，则n/i一定也是n的因子，这样保证不会超时

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
LL n,k;
int main()
{
    while(~scanf("%lld%lld",&n,&k))
    {
        vector <LL> v1,v2;
        for(LL i=1;i*i<=n;i++)
        {
            if(n%i==0)
            {
                v1.push_back(i);
                if(i*i!=n)
                   v2.push_back(n/i);
            }
        }
        LL len=v1.size()+v2.size(),len1=v1.size(),len2=v2.size();
        if(k>len)
            printf("-1\n");
        else
        {
            if(k<=len1)
                cout<<v1[k-1]<<endl;
            else
                cout<<v2[len2-(k-len1)]<<endl;
        }
    }
    return 0;
}