Codeforces Educational Codeforces Round 17 Problem.A kth-divisor (暴力+stl)

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.

solution

原本想了个分解质因数后递推，但发现空间不够

其实可以用时间换空间，暴力嘛

随便做一做就行了

#include <math.h>
#include <vector>
#include <stdio.h>
#define L long long
char B[101],*p=B,pb[101];
inline void Rin(register L &x){
    x=0;
    while(*p<'0'||*p>'9')p++;
    while(*p>='0'&&*p<='9')
        x=x*10LL+*p++-'0';
}
inline void Mo(register L x){
    register int top=0;
    while(x)pb[++top]=(x%10LL)+'0',x/=10LL;
    while(top)putchar(pb[top--]);
    putchar('\n');
}
L n,K,s,l1,l2;
std::vector<L>p1,p2;
int main(){
    fread(B,1,101,stdin);
    Rin(n),Rin(K);
    L s=sqrt(n);
    for(register L i=1;i<s;i++)
        if(n%i==0LL){
            p1.push_back(i);
            p2.push_back(n/i);
        }
    if(s*s==n)
        p1.push_back(s);
    else
        if(n%s==0){
            p1.push_back(s);
            p2.push_back(n/s);
        }
    l1=p1.size(),l2=p2.size();
    if(l1+l2<K)
        puts("-1");
    else{
        if(K<=l1)
            Mo(p1[K-1]);
        else
            Mo(p2[l2-K+l1]);
    }
    return 0;
}