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;
bool sBs(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long SUMD(long long n);
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
             long long ser);
long long MI(long long a, long long m);
void Sieve(int n = 150);
long long Powb(long long b, long long n, long long m);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int ct = log10(n) + 1;
    string s = to_string(n);
    vector<int> ans;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] != '0') {
        int k = (s[i] - '0') * pow(10, ct - i - 1);
        ans.push_back(k);
      }
    }
    cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}
long long Powb(long long b, long long n, long long m) {
  if (n == 0) return 1LL;
  if (n == 1) return b;
  long long temp = Powb(b, n / 2, m);
  if (n % 2 == 0) {
    return (temp * temp) % m;
  } else {
    return (b * ((temp * temp) % m)) % m;
  }
}
long long SUMD(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
long long BS(vector<pair<long long, long long> > &PS, long long s, long long e,
             long long ser) {
  if (s > e) return s;
  long long mid = (s + e) / 2;
  if (PS[mid].first == ser) {
    return mid;
  } else if (PS[mid].first > ser) {
    return BS(PS, s, mid - 1, ser);
  } else
    return BS(PS, mid + 1, e, ser);
}
long long MI(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}