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;
int main() {
  int n;
  scanf("%d", &n);
  int ann[1000];
  for (int i = 1; i <= n; i++) {
    int cc = 0;
    int k;
    scanf("%d", &k);
    int cnt = 4;
    while (k) {
      int ans = 1;
      for (int j = 1; j < cnt; j++) ans *= 10;
      if (k / ans) {
        ann[++cc] = k / ans * ans;
      }
      cnt--;
      k = k % ans;
    }
    printf("%d\n", cc);
    while (cc--) {
      printf("%d%c", ann[cc + 1], cc == 0 ? '\n' : ' ');
    }
    memset(ann, 0, sizeof(ann));
  }
  return 0;
}