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;
vector<long long int> isprime;
void seive(long long int size) {
  isprime.resize(size + 1, 1);
  long long int i, j;
  isprime[0] = 0;
  isprime[1] = 0;
  for (i = 2; i * i <= size; i++) {
    if (isprime[i]) {
      for (j = 2; j * i <= size; j++) isprime[i * j] = 0;
    }
  }
}
long long int power(long long int x, unsigned long long int y,
                    long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m);
  p = (p * p) % m;
  if (y % 2 != 0) p = (p * x) % m;
  return p;
}
long long int modinverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int findans() {
  long long int n;
  cin >> n;
  long long int ans = 0, i;
  vector<long long int> v;
  while (n > 0) {
    v.push_back(n % 10);
    if (n % 10 != 0) ans++;
    n /= 10;
  }
  cout << ans << endl;
  for (i = 0; i < v.size(); i++) {
    if (v[i] != 0) {
      cout << v[i];
      long long int k = i;
      while (k > 0) {
        cout << 0;
        k--;
      }
      cout << " ";
    }
  }
  cout << endl;
  return 0;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, i;
  cin >> t;
  for (i = 0; i < t; i++) {
    findans();
  }
  return 0;
}