Codeforces 762A-k-th divisor

k-th divisor

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：求一个数第k大的因子，不存在第k大的因子则输出-1

解题思路：因为数据较大，用两个vector来存因子

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longint main(){    LL n,k;    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) printf("%lld\n",v1[k-1]);            else printf("%lld\n",v2[len2-(k-len1)]);        }    }    return 0;}