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 print(vector<long long int> v) {
  for (auto ele : v) cout << ele << " ";
  cout << endl;
}
void print(long long int x) { cout << x << endl; }
void print(long long int x, long long int y) { cout << x << ", " << y << endl; }
int main() {
  long long int tcases;
  cin >> tcases;
  while (tcases--) {
    long long int number;
    cin >> number;
    long long int n = number, p = 10;
    vector<long long int> ans;
    while (n) {
      ans.push_back(n % p);
      n /= p;
    }
    long long int f = 1;
    for (long long int i = 0; i < ans.size(); i++) {
      ans[i] *= f;
      f *= 10;
    }
    vector<long long int> v;
    long long int cnt = 0;
    for (auto ele : ans) {
      if (ele) {
        v.push_back(ele);
        cnt++;
      }
    }
    print(cnt);
    print(v);
  }
}