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;
const long long INF = 1000000000000000;
const long long MOD = 998244353;
void needForSpeed() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool isPrime(long long a) {
  if (a < 2) return 0;
  for (long long i = 2; i <= sqrt(a); i++)
    if (a % i == 0) return 0;
  return 1;
}
int solve() {
  string s;
  cin >> s;
  vector<int> ans;
  for (int i = s.length() - 1; i > -1; i--) {
    if (s[i] != '0') ans.push_back((s[i] - '0') * pow(10, s.length() - 1 - i));
  }
  cout << ans.size() << '\n';
  for (int v : ans) cout << v << " ";
  cout << '\n';
  return 0;
}
int main() {
  needForSpeed();
  int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}