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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string str;
int len_of_a;
int possible_answer(int a) {
  str = to_string(a);
  len_of_a = str.length();
  int ans = 0;
  for (int i = 0; i < len_of_a; i++) {
    if (str[i] != '0') {
      ans++;
    }
  }
  return ans;
}
int main() {
  int test_case;
  scanf("%d", &test_case);
  int arr[test_case];
  for (int i = 0; i < test_case; i++) {
    scanf("%d\n", &arr[i]);
    if (arr[i] <= 10) {
      printf("1\n");
      printf("%d\n", arr[i]);
    } else if (arr[i] > 10) {
      int ans = possible_answer(arr[i]);
      printf("%d\n", ans);
      for (int j = len_of_a - 1; j >= 0; j--) {
        if (str[j] != '0') {
          printf("%c", str[j]);
          int temp = j;
          while (temp < len_of_a - 1) {
            printf("0");
            temp++;
          }
          printf(" ");
        }
      }
      printf("\n");
    }
  }
}