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 SumOfRoundNumbers{
	public static void main(String []args){
		
		java.util.Scanner scan=new java.util.Scanner(System.in);
		int n=scan.nextInt();
		int []arr=new int[n];
		for(int i=1;i<=arr.length;i++){
			arr[i-1]=scan.nextInt();
		}
		int a,b,num,x;
		for(int i=1;i<=arr.length;i++){
			num=arr[i-1];
			
			ArrayList<Integer> list=new ArrayList<Integer>();
			a=0;
			while(num!=0){
				b=num%10;
				if(b!=0){
					x=(int)Math.pow(10,a);
					list.add(b*x);
				}
				a++;
				num/=10;
			}
			System.out.println(list.size());
			for(int j=1;j<=list.size();j++){
				System.out.print(list.get(j-1)+" ");
			}
			System.out.println();
		}
	}
}