Problem: 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 (in JAVA):

Solution:
import java.util.*;

public class KthDivisor {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		long n = in.nextLong();
		long k = in.nextLong();
		
		int numDivLessSqrt = 0;
		boolean isSqrt = false;
		
		for (long p = 1; p <= n / p; p++) {
			if (n % p == 0) {
				numDivLessSqrt++;
			}
			if (p * p == n) {
				isSqrt = true;
			}
		}
		
		long ans = -1;
		if (k <= numDivLessSqrt) {
			int cnt = 0;
			for (long p = 1; p <= n / p; p++) {
				if (n % p == 0) {
					cnt++;
					if (cnt == k) {
						ans = p;
						break;
					}
				}
			}
		} else {
			if (isSqrt) {
				int cnt = 0;
				for (long p = 1; p <= n / p; p++) {
					if (n % p == 0) {
						cnt++;
						if (cnt == 2 * numDivLessSqrt - k) {
							ans = n / p;
							break;
						}
					}
				}
			} else {
				int cnt = 0;
				for (long p = 1; p <= n / p; p++) {
					if (n % p == 0) {
						cnt++;
						if (cnt == 2 * numDivLessSqrt - k + 1) {
							ans = n / p;
							break;
						}
					}
				}
			}
		}
		System.out.println(ans);
		
		in.close();
		System.exit(0);
	}
}