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;
void cf_1() {
  int t;
  cin >> t;
  while (t--) {
    long long int a, b, m, x, y, z, cnt = 0;
    cin >> a >> b;
    m = abs(a - b);
    if (a == b)
      cout << 0 << " " << 0 << "\n";
    else {
      cout << m << " ";
      x = min(b % m, m - b % m);
      cout << x << "\n";
    }
  }
}
void cf_2() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, s = 0;
    cin >> n;
    long long int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      s += a[i];
    }
    long long int j = s % n;
    long long int x = j * (n - j);
    cout << x << "\n";
  }
}
void cf_3() {
  int t;
  cin >> t;
  while (t--) {
    std::vector<int> v;
    int n, i = 1;
    cin >> n;
    while (n > 0) {
      int m = n % 10;
      if (m != 0) v.push_back(m * i);
      n /= 10;
      i *= 10;
    }
    cout << v.size() << "\n";
    for (auto &i : v) cout << i << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cf_3();
  return 0;
}