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 c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isPrime(long long n) {
  if (n < 2)
    return false;
  else if (n == 2)
    return true;
  if (n % 2 == 0) return false;
  for (int i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  c_p_c();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int temp;
    temp = n;
    int rem, cnt = 0;
    int noOfDigits = 0;
    while (temp > 0) {
      rem = temp % 10;
      if (rem != 0) cnt++;
      temp /= 10;
      noOfDigits++;
    }
    cout << cnt << endl;
    temp = n;
    for (int i = 0; i < noOfDigits; i++) {
      if (temp % 10 != 0) {
        cout << (temp % 10) * pow(10, i) << " ";
      }
      temp /= 10;
    }
    cout << endl;
  }
  return 0;
}