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.*;


import java.io.*;
public class Solution {
	static FastScanner scr=new FastScanner();
//	static Scanner scr=new Scanner(System.in);
	static PrintStream out=new PrintStream(System.out);
	static StringBuilder sb=new StringBuilder();
	
	
	
	
	static class pair{
		int x;int y;
		pair(int x,int y){ 
			this.x=x;this.y=y;
			
		}
	}
	static class triplet{
		int x;
		long y;
		int z;
		triplet(int x,long y,int z){
			this.x=x;
			this.y=y;
			this.z=z;
		}
	}
	static class FastScanner {
		 BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		} long gcd(long a,long b){
			if(b==0) {
				return a;
			}
				
			return gcd(b,a%b);
		}
	 	int[]  sort(int a[]) {
			 
	        int multiplier = 1, len = a.length, max = Integer.MIN_VALUE;
	        int b[] = new int[len];
	        int bucket[];
	        for (int i = 0; i < len; i++) if (max < a[i]) max = a[i];
	        while (max / multiplier > 0) {
	            bucket = new int[10];
	            for (int i = 0; i < len; i++) bucket[(a[i] / multiplier) % 10]++;
	            for (int i = 1; i < 10; i++) bucket[i] += (bucket[i - 1]);
	            for (int i = len - 1; i >= 0; i--) b[--bucket[(a[i] / multiplier) % 10]] = a[i];
	            for (int i = 0; i < len; i++) a[i] = b[i];
	            multiplier *= 10;
	        }
	        return a;
	    }
	 	 long modPow(long base,long exp) {
			if(exp==0) {
				return 1;
			}
			if(exp%2==0) {
				long res=(modPow(base,exp/2));
				return (res*res);
			}
			return (base*modPow(base,exp-1));
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	static void print(Object ele) {
		sb.append(ele).append(" ");
	}
	static void println(Object ele) {
		sb.append(ele).append('\n');
		
	}
	static void println() {
		sb.append('\n');
	}
//	static long LCM(long a,long b) {
//		return a*b/(scr.gcd(a, b));
//	}
//	static long n, m, k, l, q, count, sum, temp, ans;
//	static long []a;
//	static long b[];
//	static  char c[];
//	static char t[];
//	static boolean visited[][];
	static void solve() {
		long n = scr.nextLong();
		int k  = scr.nextInt();
		
		PriorityQueue<Long>pq=new PriorityQueue<>(Collections.reverseOrder());
		for(long i = 1;i<=Math.sqrt(n);i++) {
			if(n%i==0){
				pq.add(n/i);
				if((n/i)!=i) {
					pq.add(i);
				}
			}
			while(pq.size()>k) {
				pq.poll();
			}
		}
		if(pq.size()!=k) {
			out.print(-1);
			return;
		}
		out.println(pq.peek());
	}
	static int  MAX = Integer.MAX_VALUE;
	static int  MIN = Integer.MIN_VALUE;
	public static void main(String []args) {
//		int t=scr.nextInt();
//		while(t-->0) {
//			solve();
//		}
		solve();
	}
}