Problem: A positive (strictly greater than zero) integer is called round if it is of the form d00...0. In other words, a positive integer is round if all its digits except the leftmost (most significant) are equal to zero. In particular, all numbers from 1 to 9 (inclusive) are round.

For example, the following numbers are round: 4000, 1, 9, 800, 90. The following numbers are not round: 110, 707, 222, 1001.

You are given a positive integer n (1 ≤ n ≤ 10^4). Represent the number n as a sum of round numbers using the minimum number of summands (addends). In other words, you need to represent the given number n as a sum of the least number of terms, each of which is a round number.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

Each test case is a line containing an integer n (1 ≤ n ≤ 10^4).

Output

Print t answers to the test cases. Each answer must begin with an integer k — the minimum number of summands. Next, k terms must follow, each of which is a round number, and their sum is n. The terms can be printed in any order. If there are several answers, print any of them.

Example

Input


5
5009
7
9876
10000
10


Output


2
5000 9
1
7 
4
800 70 6 9000 
1
10000 
1
10 

Solution (in JAVA):

Solution:
import java.util.*;
import java.util.concurrent.ScheduledExecutorService;
import java.io.*;
import java.lang.*;
import java.math.*;
public class CodeChef
{
	public static void main(String[] args) throws IOException
	{
	   Scanner scan = new Scanner(System.in);
	   int testcases = scan.nextInt();
	   for(int m=0; m<testcases; m++)
	   {
		   int n = scan.nextInt();
		   int i=0;
		   int cnt=0;
		   StringBuffer s= new StringBuffer();
		   while(n!=0)
		   {
			   int rem=n%10;
			   n = (n-rem)/10;
			   int result = rem*(int)Math.pow(10, i);
			   if(result!=0) { cnt++; s.append(result+" "); }
			   i++;
		   }
		   System.out.println(cnt);
		   System.out.println(s);
	   }
	}
}