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 long

int 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;
}