Problem: You are given an integer n (n > 1).

Your task is to find a sequence of integers a_1, a_2, …, a_k such that:

  * each a_i is strictly greater than 1; 
  * a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n); 
  * a_{i + 1} is divisible by a_i for each i from 1 to k-1; 
  * k is the maximum possible (i. e. the length of this sequence is the maximum possible). 



If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).

It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).

Output

For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.

If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.

Example

Input


4
2
360
4999999937
4998207083


Output


1
2 
3
2 2 90 
1
4999999937 
1
4998207083 

Solution (in JAVA):

Solution:
import java.io.DataInputStream; 
import java.io.FileInputStream; 
import java.io.IOException; 
import java.io.*; 
import java.util.*; 
import java.util.StringTokenizer; 

public class Main 
{ 
	static class Reader 
	{ 
		final private int BUFFER_SIZE = 1 << 16; 
		private DataInputStream din; 
		private byte[] buffer; 
		private int bufferPointer, bytesRead; 

		public Reader() 
		{ 
			din = new DataInputStream(System.in); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public Reader(String file_name) throws IOException 
		{ 
			din = new DataInputStream(new FileInputStream(file_name)); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 

		public String readLine() throws IOException 
		{ 
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c; 
			while ((c = read()) != -1) 
			{ 
				if (c == '\n') 
					break; 
				buf[cnt++] = (byte) c; 
			} 
			return new String(buf, 0, cnt); 
		} 

		public int nextInt() throws IOException 
		{ 
			int ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do
			{ 
				ret = ret * 10 + c - '0'; 
			} while ((c = read()) >= '0' && c <= '9'); 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		public long nextLong() throws IOException 
		{ 
			long ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
			if (neg) 
				return -ret; 
			return ret; 
		} 

		public double nextDouble() throws IOException 
		{ 
			double ret = 0, div = 1; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 

			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 

			if (c == '.') 
			{ 
				while ((c = read()) >= '0' && c <= '9') 
				{ 
					ret += (c - '0') / (div *= 10); 
				} 
			} 

			if (neg) 
				return -ret; 
			return ret; 
		} 

		private void fillBuffer() throws IOException 
		{ 
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
			if (bytesRead == -1) 
				buffer[0] = -1; 
		} 

		private byte read() throws IOException 
		{ 
			if (bufferPointer == bytesRead) 
				fillBuffer(); 
			return buffer[bufferPointer++]; 
		} 

		public void close() throws IOException 
		{ 
			if (din == null) 
				return; 
			din.close(); 
		} 
	} 

	public static void main(String[] args) throws IOException 
	{ 
		Reader in=new Reader(); 
		PrintWriter pw = new PrintWriter(System.out);
	    int t=in.nextInt();
	    while(t>0){
	        long n=in.nextLong();
	        List<Long> li=primeFactors(n);
	        Map<Long,Long> freq=new HashMap<>();
	        long q=1;
	        for(int i=0;i<li.size();i++){
	        if(freq.containsKey(li.get(i))){
			    long x=freq.get(li.get(i));
			    x++;
			    freq.replace(li.get(i),x);
			}else freq.put(li.get(i),q);
	        }
	        long key=-1,value=-1;
	        for (Map.Entry<Long,Long> m : freq.entrySet()){
	            if(m.getValue()>value){
	                key=m.getKey();
	                value=m.getValue();
	            }
	        }
	        Collections.sort(li);
	        pw.println(value);
	        for(long i=1;i<=value-1;i++){
	           pw.print(key+" "); 
	           li.remove(new Long(key));
	        } 
	        long prod=1;
	        for(int i=0;i<li.size();i++){
	            prod*=li.get(i);
	        }
	        pw.print(prod);
	        pw.println();
	        pw.flush();
	        t--;
	    }
	} 

	public static List<Long> primeFactors(long n) 
	{ 
		// Print the number of 2s that divide n 
		List<Long> li=new ArrayList<>();
		
		while (n%2==0) 
		{ 
			li.add(2l);
			
			n /= 2; 
		} 

		// n must be odd at this point. So we can 
		// skip one element (Note i = i +2) 
		for (long i = 3; i*i <= n; i+= 2) 
		{ 
			// While i divides n, print i and divide n 
			while (n%i == 0) 
			{ 
				li.add(i); 
				n /= i; 
			} 
		} 

		// This condition is to handle the case whien 
		// n is a prime number greater than 2 
		if (n > 2) li.add(n);
		return li;
	} 

}