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.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static PrintWriter out;
    static int  n, m, maxn = 1;
    static String ss;
    static LinkedList<Integer> linkedList = new LinkedList<>();
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = Integer.parseInt(br.readLine().trim());
        while(t-->0){
            String temp = br.readLine().trim();
            n = Integer.parseInt(temp);
            int idx = 0;
            int[] p = new int[10];
            for (int i = temp.length()-1, j = 0; i >= 0; i--, j++) {
                if(temp.charAt(i) >= '1' && temp.charAt(i) <= '9'){
                    p[idx++] = (int) ((int)(temp.charAt(i)-'0')*Math.pow(10,j));
                }
            }
            out.println(idx);
            for(int i = 0; i < idx ; i++){
                out.print(p[i]+" ");
            }
            out.println();
        }


        out.close();
    }



    static class MyUtils {
        //《---------------------------工具类---------------------------》\

        //从大到小排序
        static Comparator<Integer> cmp = new Comparator<Integer>() {
            public int compare(Integer i1, Integer i2) {
                return i2-i1;
            }
        };

        //[begin, end) 进行二分查找，返回 大于或等于 tar的第一个元素位置。如果所有元素都小于tar，则返回 end.
        public static int lowerBound(int []nums,int l,int r,int target){
            while(l<r){
                int m = (l+r)/2;
                if(nums[m]<=target) l = m+1;
                else    r = m;
            }
            return l;
        }
        //全排列
        public static void perm(int[] array,int start,int end) {
            if(start==end) {
                System.out.println(Arrays.toString(array));
            } else {
                for (int i = start; i <= end; i++) {
                    //1，2，3的全排列这块相当于将其中一个提了出来，下次递归从start+1开始
                    swap(array,start,i);
                    perm(array,start+1,end);
                    //这块是复原数组，为了保证下次另外的同级递归使用数组不会出错
                    //这块可以通过树来理解，每次回退一步操作，交换回去
                    swap(array,start,i);
                }
            }
        }
        //交换两数
        public static void swap(int[] array,int i,int j) {
            int temp = array[i];
            array[i] = array[j];
            array[j] = temp;
        }

        private static int gcd(int a, int b) {
            int t ;
            if(a < b){
                t=a;
                a=b;
                b=t;
            }
            return a%b==0?b:(gcd(b,a%b));
        }

        // [begin, end) 进行二分查找，返回 大于 tar的第一个元素位置。如果所有元素都小于tar，则返回 end.
        public static int upperBound(int []nums ,int l,int r, int target){
            while(l<r){
                int m = (l+r)/2;
                if(nums[m]>=target) r= m;
                else    l = m +1;
            }
            return l;
        }
    }

    static class Pairs {
        int first;
        int second;
        public Pairs(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
}