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;
long long a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w,
    x, y, z;
long long ans = 0, sum = 0, mn = 1e18, mx = 0, cnt = 0;
const long long M = 1000000007;
map<int, int> mp;
vector<int> vec;
vector<int> vec2;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  cin >> t;
  while (t--) {
    string n;
    cin >> n;
    cnt = 0;
    for (int i = 0; i < n.size(); i++) {
      if (n[i] != '0') cnt++;
    }
    cout << cnt << endl;
    for (int i = 0; i < n.size(); i++) {
      if (n[i] != '0') {
        cout << n[i];
        for (int j = 0; j < n.size() - i - 1; j++) {
          cout << '0';
        }
        cout << ' ';
      }
    }
    cout << endl;
  }
  return 0;
}