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 Main{
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);  
      int t = sc.nextInt();
      while(t-- != 0){
          int n = sc.nextInt();
          int n1 = n;
          int i = 0,c = 0;
          while(n!=0){
              int rem = n%10;
              if(rem != 0)  c++;
              n = n/10;
          }
          System.out.println(c);
          while(n1!=0){
              int rem = n1%10;
              if(rem != 0)  System.out.print(rem * (int)Math.pow(10,i)+" ");
              i++;
              n1 = n1/10;
          }
          System.out.println();
      }
    }
}