CodeForces - 762A k-th divisor (数学

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本文介绍了一种算法，用于解决给定两个整数n和k时如何找到n的第k个最小因数的问题。如果不存在这样的因数，则输出-1。该算法通过遍历并检查每个可能的因数来确定答案。

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

Input

5 3

Output

-1

Input

12 5

Output

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：给两个数n,k，求第一个数的第k个因数
如果没有这输出-1
~~蒟蒻的博主会说题意看错了cf成功爆零么~~

#include<stdio.h>
#include<math.h>
#define LL long long

int main()
{
    LL n, k;
    while(~scanf("%lld%lld",&n,&k)) {
        LL x = sqrt(n);
        bool flag = false;
        LL ans_1 = 0, ans = 0;
        for(LL i = 1;i <= x; i++) {
            if(n%i == 0) 
                ans++;
                if(ans == k) {
                printf("%lld\n",i);
                flag = true;
                break;
            }
        }
        if(!flag){
                if(x*x == n) {
            ans = ans*2 - 1;
        }
        else {
            ans = ans*2; 
        }
        if(ans < k) {
            printf("-1\n");
            flag = true;
        }
        }
        if(!flag) {
        ans = ans + 1 -k;
        for (LL i = 1;i <= x; i++) {
            if(n%i == 0) {
                ans_1++;
                if(ans_1 == ans) {
                    printf("%lld\n",(LL)n/i);
                    break;
                }
            }
        }
        }

    }
return 0;
}