Problem: To confuse the opponents, the Galactic Empire represents fractions in an unusual format. The fractions are represented as two sets of integers. The product of numbers from the first set gives the fraction numerator, the product of numbers from the second set gives the fraction denominator. However, it turned out that the programs that work with fractions in this representations aren't complete, they lack supporting the operation of reducing fractions. Implement this operation and the Empire won't forget you.

Input

The first input line contains two space-separated integers n, m (1 ≤ n, m ≤ 105) that show how many numbers the first set (the numerator) and the second set (the denominator) contain, correspondingly.

The second line contains n space-separated integers: a1, a2, ..., an (1 ≤ ai ≤ 107) — the numbers that are multiplied to produce the numerator.

The third line contains m space-separated integers: b1, b2, ..., bm (1 ≤ bi ≤ 107) — the numbers that are multiplied to produce the denominator.

Output

Print the answer to the problem in the form, similar to the form of the input data. The number of values in the sets you print nout, mout must satisfy the inequality 1 ≤ nout, mout ≤ 105, and the actual values in the sets aout, i and bout, i must satisfy the inequality 1 ≤ aout, i, bout, i ≤ 107. 

Separate the values in the lines by spaces. The printed fraction must be reduced, that is, there mustn't be such integer x (x > 1), that the numerator and the denominator of the printed fraction are divisible by x. If there are several matching answers, print any of them.

Examples

Input

3 2
100 5 2
50 10


Output

2 3
2 1
1 1 1


Input

4 3
2 5 10 20
100 1 3


Output

1 1
20
3

Note

In the first test sample the numerator equals 1000, the denominator equals 500. If we reduce fraction 1000/500 by the greatest common divisor of the numerator and the denominator (by 500), we obtain fraction 2/1.

In the second test sample the numerator equals 2000, the denominator equals 300. If we reduce fraction 2000/300 by the greatest common divisor of the numerator and the denominator (by 100), we obtain fraction 20/3.

Solution (in JAVA):

Solution:


import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;


@SuppressWarnings("unused")
public class Reducing222C {

	InputStream is;
	PrintWriter out;
	String INPUT = "";

	int mod = (int)(Math.pow(10,9)+7);

	ArrayList<Integer> al = new ArrayList<>();

	int spf[], z = (int)(1e7 + 123);

	int[] play(int[] a, int n) {
		int d[] = new int[z];
		for(int x : a) {
			while(spf[x] > 1) {
				d[spf[x]]++;
				x /= spf[x];
			}
		}
		return d;
	}

	int[] game(int[] a, int[] d, int n) {
		for(int i = 0; i < n; i++) {
			int x = a[i];
			while(spf[x] > 1) {
				int u = spf[x];
				if(d[u] > 0) {
					a[i] /= u;  d[u]--;
				}
				x /= u;
			}
		}
		return a;
	}
	
	void solve() {

		spf = new int[z];
		for(int i = 0; i < z; i++)  spf[i] = i;
		for(int i = 2; i*i <= z; i++) {
			if(spf[i] == i) {
				for(int j = i*i; j < z; j += i) {
					if(spf[j] == j) spf[j] = i;
				}
			}
		}
		int n = ni(), m = ni();
		int a[] = new int[n];
		int b[] = new int[m];
		for(int i = 0; i < n; i++)  a[i] = ni();
		for(int i = 0; i < m; i++)  b[i] = ni();
		int ap[] = play(a, n);
		int bp[] = play(b, m);
		for(int i = 0; i < z; i++) {
			ap[i] = Math.min(ap[i], bp[i]);
			bp[i] = ap[i];
		}
		a = game(a, ap, n);
		b = game(b, bp, m);
		out.println(n +" "+ m);
		for(int x : a)  out.print(x +" ");
		out.println();
		for(int x : b)  out.print(x +" ");
		out.println();
	}
	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
		//tr(System.currentTimeMillis()-s+"ms");
	}

	public static void main(String[] args) throws Exception { new Reducing222C().run(); }

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }

	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}

	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}

	private int[][] na(int n , int m)
	{
		int[][] a = new int[n][m];
		for(int i = 0;i < n;i++) 
			for(int j = 0 ; j<m ; j++) a[i][j] = ni();
		return a;
	}

	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	void display2D(int a[][]) {
		for(int i[] : a) {
			for(int j : i) {
				out.print(j+" ");
			}
			out.println();
		}
	}

	int[][] creategraph(int n , int m) {
		int g[][] = new int[n+1][];

		int from[] = new int[m];
		int to[] = new int[m];
		int ct[] = new int[n+1];

		for(int i = 0 ; i<m; i++) {
			from[i] = ni();
			to[i] = ni();
			ct[from[i]]++;
			ct[to[i]]++;
		}

		int parent[] = new int[n+1];

		for(int i = 0 ; i<n+1 ; i++) g[i] = new int[ct[i]];

		for(int i = 0 ; i<m ; i++) {
			g[from[i]][--ct[from[i]]] = to[i];
			g[to[i]][--ct[to[i]]] = from[i];
		}

		return g;
	}

	static long __gcd(long a, long b)
	{

		if(b == 0) 
		{
			return a;
		}
		else
		{
			return __gcd(b, a % b);
		}
	}

	// To compute x^y under modulo m
	static long power(long x, long y, long p)
	{

		// Initialize result
		long res = 1;

		// Update x if it is more than or
		// equal to p
		x = x % p;

		while (y > 0)
		{

			// If y is odd, multiply x
			// with result
			if (y % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
		x = (x * x) % p;
		}

		return res;
	}

	// Function to find modular 
	// inverse of a under modulo m
	// Assumption: m is prime
	static long modInverse(long a, int m)
	{
		if (__gcd(a, m) != 1) {
			//System.out.print("Inverse doesn't exist");
			return -1;
		}

		else {

			// If a and m are relatively prime, then
			// modulo inverse is a^(m-2) mode m
			//			System.out.println("Modular multiplicative inverse is "
			//											+power(a, m - 2, m));
			return power(a, m - 2, m);
		}
	}

	static long nCrModPFermat(int n, int r,
			int p , long fac[])
	{

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long t = (fac[n]* modInverse(fac[r], p))%p;

		return ( (t* modInverse(fac[n-r], p))% p);
	}


	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}