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 INF = (int)1E9 + 5;
const long long LINF = (long long)4E18 + 5;
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? abs(a) : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    vector<int> v;
    int count = 0, pcount = 0;
    cin >> n;
    if (n < 10) {
      cout << 1 << '\n' << n << '\n';
    } else {
      while (n != 0) {
        if (n % 10 != 0) {
          v.push_back((n % 10) * (int)(pow(10, pcount) + 0.5));
          count++;
        }
        n /= 10;
        pcount++;
      }
      cout << count << '\n';
      for (int i = (0); i < (count); i++) cout << v[i] << ' ';
      cout << '\n';
    }
  }
  return 0;
}