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;
using ll = long long;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
vector<long long> fact;
ll ncr2d[150][150];
ll dp[105][10005];
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
ll inverse_power(ll a) { return binpow(a, 1000000007 - 2); }
ll nCr(ll n, ll r) {
  if (r > n) return 0;
  return ((fact[n] * inverse_power(fact[r])) % 1000000007 *
          inverse_power(fact[n - r])) %
         1000000007;
}
void factorial() {
  fact.push_back(1);
  for (int i = 2; i < 100000; i++) {
    fact.push_back((fact.back() * i) % 1000000007);
  }
}
int find_element(vector<string> &vec, string search) {
  for (int i = 0; i < vec.size(); i++) {
    if (vec[i] == search) {
      return i;
    }
  }
  return -1;
}
void solve() {
  int n;
  cin >> n;
  vector<int> arr;
  int i = 1;
  while (n > 0) {
    if (n % 10 != 0) {
      arr.push_back((n % 10) * i);
    }
    n = n / 10;
    i = i * 10;
  }
  cout << arr.size() << '\n';
  for (auto item : arr) {
    cout << item << " ";
  }
  cout << '\n';
}
int main() {
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}