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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
void solve() {
  int n;
  cin >> n;
  int pw = 1;
  vector<int> v;
  while (n) {
    if (n % 10) v.emplace_back((n % 10) * pw);
    n /= 10;
    pw *= 10;
  }
  cout << (int)v.size() << "\n";
  for (auto i : v) cout << i << " ";
  cout << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  for (__typeof(t) i = (0) - ((0) > (t)); i != (t) - ((0) > (t));
       i += 1 - 2 * ((0) > (t))) {
    solve();
  }
  return 0;
}