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 t;
  cin >> t;
  while (t--) {
    int n;
    int x[5];
    int k = 0;
    cin >> n;
    string nstr = to_string(n);
    if (nstr.size() == 1) {
      k = 1;
      cout << k << " " << endl << n << endl;
    } else {
      x[0] = n % 10;
      n = n - x[0];
      x[1] = n % 100;
      n = n - x[1];
      x[2] = n % 1000;
      n = n - x[2];
      x[3] = n % 10000;
      n = n - x[3];
      x[4] = n % 100000;
      n = n - x[4];
      vector<int> num = {x[0], x[1], x[2], x[3], x[4]};
      for (auto i : num)
        if (i != 0) {
          k++;
        }
      cout << k << endl;
      for (auto i : num)
        if (i != 0) {
          cout << i << " ";
        }
      cout << endl;
    }
  }
}