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>
int power(int x, int y) {
  int i;
  int num = 1;
  for (i = 0; i < y; i++) {
    num = num * x;
  }
  return num;
}
int main() {
  int t;
  scanf("%d", &t);
  int n, i, j, k, l;
  int b[10];
  for (i = 0; i < t; i++) {
    l = 0;
    scanf("%d", &n);
    for (j = 0; j < 10; j++) {
      if (n == 0) break;
      if (n % 10 == 0) {
        n = n / 10;
        continue;
      }
      b[l] = (n % 10) * (power(10, j));
      n = n / 10;
      l++;
    }
    printf("%d \n", l);
    for (k = 0; k < l; k++) {
      printf("%d ", b[k]);
    }
    printf("\n");
  }
  return 0;
}