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 prime(long long int s) {
  if (s <= 1) return false;
  if (s <= 3) return true;
  if (s % 2 == 0 || s % 3 == 0) return false;
  for (long long int i = 5; i * i <= s; i = i + 6)
    if (s % i == 0 || s % (i + 2) == 0) return false;
  return true;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  string s;
  cin >> s;
  long long int n = s.size();
  vector<long long int> ans;
  for (long long int i = 0; i < s.size(); i++) {
    if (s[i] != '0') {
      long long int a = s[i] - '0';
      long long int b = 1;
      for (long long int j = 0; j < n - i - 1; j++) b *= 10;
      long long int c = a * b;
      ans.push_back(c);
    }
  }
  cout << ans.size() << "\n";
  for (long long int num : ans) cout << num << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}