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 
#include <bits/stdc++.h>
using namespace std;
long long const mod = 998244353;
long long const mod1 = 1000000007;
long long toint(const string &s) {
  stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
string tostring(long long number) {
  stringstream ss;
  ss << number;
  return ss.str();
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n & 1) {
    return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
  } else {
    return (modexp((x * x) % mod, n / 2) % mod);
  }
}
long long modinv(long long n) { return modexp(n, mod - 2); }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    string n;
    cin >> n;
    long long c = 0;
    vector<long long> v;
    long long s = n.length();
    for (long long i = 0; i < n.length(); i++) {
      if (n[i] != '0') {
        v.push_back((n[i] - '0') * pow(10, s - i - 1));
      }
    }
    cout << v.size() << endl;
    for (auto x : v) cout << x << " ";
    cout << endl;
  }
}
