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 
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class CodeForces640 {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		while(t>0) {
			int a=s.nextInt();
			List<Integer> l=RoundSum(a);
			int si=l.size();
			System.out.println(si);
			for(int i=0;i<si;i++) {
				System.out.println(l.get(i)+" ");
			}
			System.out.println();
			t--;
		}
	}
	public static List<Integer> RoundSum(int num) {
		int temp=num;
		int count=0;
		int digit;
		List<Integer> ll=new ArrayList<>();
//		int r=0;
	    while(num > 0){
	            num = num / 10;
	            count++;
	        }
	    int r=0;
	        while(temp > 0)
	        {
	            digit = temp % 10;
	            if(digit!=0) {
	            	ll.add((int) ((digit)*Math.pow(10,r)));
	            	r++;
	            }
	            else {r++;}
	            temp = temp / 10;
	            count--;
	        }
			return ll;
	    
	}
}