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;
template <class T>
using V = vector<T>;
using Graph = vector<vector<int>>;
using P = pair<long long, long long>;
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
int f(int n) {
  int res = 1;
  for (int i = 0; i < n; i++) res *= 10;
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    string s;
    s = to_string(n);
    reverse((s).begin(), (s).end());
    V<int> ans;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] - '0' != 0) {
        ans.push_back((s[i] - '0') * f(i));
      }
    }
    cout << ans.size() << endl;
    for (int a : ans) cout << a << ' ';
    cout << endl;
  }
}
