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:
//sum of round numbers 1352A
import java.util.Scanner;
import java.lang.Math;

public class pb1
{
	public static void main(String args[])
	{
		Scanner in = new Scanner(System.in);

		int t;
		t = in.nextInt();

		for(int y=1; y<=t; y++)
		{
			int n;
			n = in.nextInt();

			if(n>=1 && n<=9)
			{
				System.out.println(1);
				System.out.println(n);
			}
			else
			{ 
				int r[] = new int[5];
				int c=0;
				for(int i=0; i<5; i++)
				{
					r[i] = (n%10)*(int)Math.pow(10,i);
					if(r[i] != 0)
						c += 1;
					n /= 10;
					if(n == 0)
						break;
				}
				System.out.println(c);
				for(int i=0; i<5; i++)
				{
					if(r[i] != 0)
						System.out.print(r[i] + " ");
				}
				System.out.println();
			}

		}
	}
}
//500