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.*;

public class sum_of_round_numbers {
		
	public static void main(String[] args) 
	{
		Scanner sc=new Scanner(System.in);
		int tc = sc.nextInt();
		for(int i=0; i<tc; i++)
		{
			int n=sc.nextInt();
			int tmp=n, ctr=0, rev=0;
			
			while(tmp>0)
			{
				tmp/=10;
				ctr++;
			}
			int[] arr=new int[ctr];
			int k=ctr-1;
			int ctr3=0;
			while(n>0&&k>=0)
			{
				if(n%10!=0)
				{
					ctr3++;
				}
				arr[k]=n%10;
				n/=10;
				k--;
			}
			
			int ctr2=0;
			System.out.println(ctr3);
			for(int j=arr.length-1; j>=0;j--)
			{
				if(arr[j]*Math.pow(10, ctr2)>0)
				{
					System.out.print((int)(arr[j]*Math.pow(10, ctr2))+" ");
				}
				ctr2++;
			}
			System.out.println();
			
		}
		
		
	}
}