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;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
long long poww(long long a, long long b) {
  if (b == 0) return 1;
  long long x = poww(a, b / 2);
  x = x * x;
  if (b % 2 == 1) x = (x * a);
  return x;
}
int frq[200000], a[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> v;
    for (int i = 0; i < (n); i++) {
      if (s[i] != '0') {
        long long now = poww(10, n - i - 1);
        long long x = s[i] - '0';
        v.push_back(now * x);
      }
    }
    cout << v.size() << endl;
    for (auto x : v) cout << x << " ";
    cout << endl;
  }
  return 0;
}