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 
#include <bits/stdc++.h>
using namespace std;
const int MAX = (int)1e5 + 5;
string check(char x, int n) {
  string it = "";
  it.push_back(x);
  string z = "";
  for (int i = 0; i < n; i++) {
    z = z + "0";
  }
  it = it + z;
  return it;
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    int x;
    int count = 0;
    vector<string> v;
    cin >> x;
    string it = to_string(x);
    int size = it.length();
    for (int i = 0; i < size; i++) {
      if (it[size - 1 - i] != '0') {
        string second = check(it[size - 1 - i], i);
        count++;
        v.push_back(second);
      }
    }
    reverse(v.begin(), v.end());
    cout << count << endl;
    for (auto& i : v) {
      cout << i << " ";
    }
    cout << endl;
  }
  return 0;
}
