Problem: Bizon the Champion isn't just friendly, he also is a rigorous coder.

Let's define function f(a), where a is a sequence of integers. Function f(a) returns the following sequence: first all divisors of a1 go in the increasing order, then all divisors of a2 go in the increasing order, and so on till the last element of sequence a. For example, f([2, 9, 1]) = [1, 2, 1, 3, 9, 1].

Let's determine the sequence Xi, for integer i (i ≥ 0): X0 = [X] ([X] is a sequence consisting of a single number X), Xi = f(Xi - 1) (i > 0). For example, at X = 6 we get X0 = [6], X1 = [1, 2, 3, 6], X2 = [1, 1, 2, 1, 3, 1, 2, 3, 6].

Given the numbers X and k, find the sequence Xk. As the answer can be rather large, find only the first 105 elements of this sequence.

Input

A single line contains two space-separated integers — X (1 ≤ X ≤ 1012) and k (0 ≤ k ≤ 1018).

Output

Print the elements of the sequence Xk in a single line, separated by a space. If the number of elements exceeds 105, then print only the first 105 elements.

Examples

Input

6 1


Output

1 2 3 6 


Input

4 2


Output

1 1 2 1 2 4 


Input

10 3


Output

1 1 1 2 1 1 5 1 1 2 1 5 1 2 5 10 

Solution (in JAVA):

Solution:
//package codeforces;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class E {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(System.out);
    StringTokenizer stringTokenizer;

    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    void solve() throws IOException {
        long x = nextLong();
        long k = nextLong();
        List<Long> d = new ArrayList<>();
        for (long y = 1; y * y <= x; y++) {
            if (x % y == 0) {
                long z = x / y;
                d.add(y);
                if (z != y) d.add(z);
            }
        }
        final long[] divisors = new long[d.size()];
        for (int i = 0; i < d.size(); i++) {
            divisors[i] = d.get(i);
        }
        Arrays.sort(divisors);
        class Utils {
            int limit = 100000;

            void rec(long x, long k) {
                if (limit == 0) return;
                if (x == 1) {
                    writer.print("1 ");
                    limit--;
                    return;
                }
                if (k == 0) {
                    writer.print(x + " ");
                    limit--;
                    return;
                }
                for (long y : divisors) {
                    if(y > x) return;
                    if(x % y == 0) {
                        rec(y, k - 1);
                        if(limit == 0) return;
                    }
                }
            }
        }
        new Utils().rec(x, k);
        writer.close();
    }


    public static void main(String[] args) throws IOException {
        new E().solve();
    }
}