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;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b);
long long int gcd(long long int a, long long int b);
int prime(int n);
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    string a;
    cin >> a;
    long long int n = a.size();
    long long int k = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != '0') k++;
    }
    cout << k << endl;
    for (int i = 0; i < n; i++) {
      if (a[i] != '0') {
        cout << a[i];
        for (int j = i + 1; j < n; j++) cout << "0";
        cout << " ";
      }
    }
    cout << endl;
  }
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int prime(int n) {
  int flag = 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}