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.*;
import java.io.*;
public class Problem1{
    StringTokenizer st;
    BufferedReader br;
    public void run() throws Exception{
        br = new BufferedReader(new InputStreamReader(System.in));
        int t = in();
        while(t-- > 0){
            int n = in(), ans = 0, tmp = 1;
            List<Integer> nums = new ArrayList<Integer>();
            while(n != 0){
                if(n%10 != 0){
                    nums.add((n%10)*tmp);
                    ans++;
                }
                n /= 10;
                tmp *= 10;
            }
            System.out.println(ans);
            for(int num: nums)
                System.out.print(num + " ");
            System.out.print("\n");
        }

    }
    Integer in() throws Exception{
        return Integer.parseInt(nextToken());
    }
    String nextToken() throws Exception{
        if(st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }
    public static void main(String[] args) throws Exception{
        new Problem1().run();
    }
}