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;
const int maxn = 10000;
int main() {
  int a[maxn];
  int b[5];
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    memset(b, 0, sizeof(b));
    cin >> a[i];
    int cnt = 1, k = 0, j = 0;
    while (1) {
      int temp;
      if (a[i] / 10 == 0) {
        int m = pow(10, k);
        b[j++] = a[i] * m;
        break;
      }
      if (a[i] / 10 != 0 && a[i] % 10 != 0) {
        cnt++;
        int m = pow(10, k);
        temp = (a[i] % 10) * m;
        b[j++] = temp;
      }
      a[i] = a[i] / 10;
      k++;
    }
    j = 0;
    printf("%d\n", cnt);
    printf("%d", b[j++]);
    while (b[j] != 0) printf(" %d", b[j++]);
    printf("\n");
  }
  return 0;
}