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 checkPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 1) return (power(a, b - 1) * a) % 1000000007;
  long long q = power(a, b / 2);
  return (q * q) % 1000000007;
}
bool CPT(long long n) { return !(n & (n - 1)); }
long long gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }
bool isSubsetSum(int set[], int n, int sum) {
  bool subset[n + 1][sum + 1];
  for (int i = 0; i <= n; i++) subset[i][0] = true;
  for (int i = 1; i <= sum; i++) subset[0][i] = false;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= sum; j++) {
      if (j < set[i - 1]) subset[i][j] = subset[i - 1][j];
      if (j >= set[i - 1])
        subset[i][j] = subset[i - 1][j] || subset[i - 1][j - set[i - 1]];
    }
  }
  return subset[n][sum];
}
bool compare(string a, string b) {
  string ab = a + b;
  string ba = b + a;
  return ab > ba;
}
int summands(int n) {
  long long c = 0;
  while (n > 0) {
    long long rem = n % 10;
    if (rem != 0) c++;
    n /= 10;
  }
  return c;
}
void solve() {
  long long n;
  cin >> n;
  long long i = 1;
  cout << summands(n) << "\n";
  while (n > 0) {
    long long rem = n % 10;
    if (rem * i != 0) cout << rem * i << " ";
    i *= 10;
    n /= 10;
  }
  cout << "\n";
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}