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.Scanner;

public class Program {
    private static final String[] zeroes = {"", "0", "00", "000", "0000"};

    public static void main(String[] args) {
        final var sc = new Scanner(System.in);
        final var t = sc.nextInt();

        for (var i = 0; i < t; i++) {
            solve(sc.nextInt());
        }
    }

    private static void solve(int n) {
        final var digits = String.valueOf(n).toCharArray();
        var k = 0;
        for (char digit : digits) {
            if (digit != '0') {
                k++;
            }
        }
        System.out.println(k);

        for (int i = 0; i < digits.length; i++) {
            if (digits[i] != '0') {
                System.out.printf("%s%s ", digits[i], zeroes[digits.length - i - 1]);
            }
        }
        System.out.println();
    }
}