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.Scanner;
import java.util.Stack;

public class ProblemA {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner s = new Scanner(System.in);
		
		int t = s.nextInt();
		
		for(int i=0;i<t;i++) {
			int n = s.nextInt();
			Stack<Integer> stack = new Stack<>();
			
			int pow = 1;
			int count = 0;
			if(n%10 != 0)
				count++;
			stack.push(n%10);
			n = n/10;
			while(n != 0) {
				stack.push(n%10);
				if(n%10 != 0)
					count++;
				n = n/10;
				pow = pow * 10;
			}
			System.out.println(count);
			while(!stack.isEmpty()) {
				int v = stack.pop();
				if(v != 0)
					System.out.print(v*pow+" ");
				
				pow = pow/10;
			}
			System.out.println();
		}

	}

}