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.lang.*;

public class Rextester
{  
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t;
        t=sc.nextInt();
        while(t>0)
        {
            int m;
            m=sc.nextInt();
            int temp[]=new int[5];
            int ct=0,n=m,k=0,i;
            if(n/10000>=1)
            {
                
                ct++;
                temp[k++]=(n/10000)*10000;
                n=n-temp[k-1];
            }
             if(n/1000>=1)
            {
                
                ct++;
                temp[k++]=(n/1000)*1000;
                n=n-temp[k-1];
            }
            if(n/100>=1)
            {
                
                ct++;
                temp[k++]=(n/100)*100;
                n=n-temp[k-1];
            }
            if(n/10>=1)
            {
                
                ct++;
                temp[k++]=(n/10)*10;
                n=n-temp[k-1];
            }
            if(n>0){
            temp[k]=n;
                ct++;}
            System.out.println(ct);
            for(i=0;i<ct;i++)
                System.out.print(temp[i]+" ");
            t=t-1;
            System.out.println();
        }  
    }
}