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;
using ll = long long;
using ld = long double;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} fast_ios_;
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < (int)(t); ++i) {
    int n;
    cin >> n;
    int d = 0;
    vector<int> ls;
    while (n != 0) {
      int a = n % 10;
      if (a != 0) {
        ls.push_back(a * pow(10, d));
      }
      n /= 10;
      d++;
    }
    cout << ls.size() << endl;
    reverse(ls.begin(), ls.end());
    for (int j = 0; j < (int)(ls.size()); ++j) {
      cout << ls[j];
      if (j - 1 != ls.size()) {
        cout << " ";
      }
    }
    cout << endl;
  }
  return 0;
}