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.HashSet;
import java.util.Scanner;

public class SumOfRoundNumbers {

	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int t=scan.nextInt();
		while(t-->0) {
			int n=scan.nextInt();
			HashSet<Integer> hs=new HashSet<Integer>();
			int pow=1;
			while(n>0) {
				if(n%10>0) {
				 int a=n%10;
				  a=a*pow;
				  hs.add(a);
				}
				n=n/10;;
				pow*=10;
			}
			Object[] a=hs.toArray();
			System.out.println(hs.size());
			for(int i=0;i<hs.size();i++) {
				System.out.print(a[i]+" ");
			}
			
		}

	}

}