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.math.*;

public class S{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		for(int q = 0; q < n; q++){
			String s = in.next();
			int cnt = 0;
			for(int i = 0; i < s.length(); i++){
				if(s.charAt(i) == '0')
					cnt++;

			}
			pl(""+ (s.length() - cnt));
			for(int i = 0; i < s.length(); i++){
				if(s.charAt(i) != '0')
					System.out.print(((s.charAt(i) - 48) *(int) Math.pow(10,(s.length() - i - 1)))+" ");
			}
			pl("");
			
		}
	}

	public static void pl(String s){
		System.out.println(s);
	}
}