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CodeForces762A

A. 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.

 1 //2017.02.01
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int divisor[35000000];
 9 
10 int main()
11 {
12     long long n, k;
13     while(cin>>n>>k)
14     {
15         long long ans = -1;
16         int cnt = 0;
17         bool fg = false;
18         for(long long i = 1; i*i <= n; i++)
19         {
20             if(n%i==0)divisor[++cnt] = i;
21             if(i*i == n)fg = true;
22         }
23         if(fg && k > 2*cnt-1)cout<<-1<<endl;
24         else if(k > 2*cnt)cout<<-1<<endl;
25         else if(k<=cnt)cout<<divisor[k]<<endl;
26         else{
27             if(!fg)cout<<(n/divisor[2*cnt+1-k])<<endl;
28             else cout<<(n/divisor[2*cnt-k])<<endl;
29         }
30     }
31 
32     return 0;
33 }

 

posted @ 2017-02-01 20:00  Penn000  阅读(276)  评论(0编辑  收藏  举报