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:
import java.util.*;
import java.io.*;
public class E {
    static HashMap<Long, List<Long>> divMap;
    static List<Long> divs;
    static long x, k;
    static void solve() {
        x = in.nextLong();
        k = in.nextLong();
        if (x == 1) {
            out.println(1);
            return;
        }
        divMap = new HashMap<>();
        divs = new ArrayList<>();
        for (long i = 1; i * i <= x; i++) {
            if (x % i == 0) {
                divs.add(i);
                if (i * i != x)
                    divs.add(x / i);
            }
        }
        Collections.sort(divs);
        for (long div : divs) {
            List<Long> divs2 = new ArrayList<>();
            for (long j = 1; j * j <= div; j++) {
                if (div % j == 0) {
                    divs2.add(j);
                    if (j * j != div)
                        divs2.add(div / j);
                }
            }
            Collections.sort(divs2);
            divMap.put(div, divs2);
        }
        // d(divs.size() + "");
        List<Long> res = new ArrayList<>();
        o(x, 0, res);
        p(res);
    }
    static void o(long cur, long depth, List<Long> res) {
        if (res.size() == 100000) return;
        if (depth == k || cur == 1) {
            res.add(cur);
            return;
        }
        for (long div : divMap.get(cur)) {
            o(div, depth + 1, res);
        }
    }
    static void p(List<Long> res) {
        for (int i = 0; i < res.size(); i++) {
            if (i != 0) out.print(" ");
            out.print(res.get(i));
        }
    }
    static PrintWriter out = new PrintWriter(System.out);
    static InputReader in = new InputReader(System.in);
    static void d(Object o) {
        out.println(o);
        out.flush();
    }
    public static void main(String args[]) {
        solve();
        out.close();
    }
}
class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;
    public InputReader(InputStream stream) {
	reader = new BufferedReader(new InputStreamReader(stream), 32768);
	tokenizer = null;
    }
    public String next() {
	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	    try {
		tokenizer = new StringTokenizer(reader.readLine());
	    } catch (IOException e) {
		throw new RuntimeException(e);
	    }
	}
	return tokenizer.nextToken();
    }
    public double nextDouble() {
	return Double.parseDouble(next());
    }
    public long nextLong() {
	return Long.parseLong(next());
    }
    public int nextInt() {
	return Integer.parseInt(next());
    }
}