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;
void ritik() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long int gcd(long long int A, long long int B) {
  if (B == 0)
    return A;
  else
    return gcd(B, A % B);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return (res % p);
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(min(a, b), c);
}
long long int nCr(long long int n, long long int r) {
  long long int x = n - r, y = 0, z = 1;
  while (y < r) {
    z = (z * (x + 1)) / (y + 1);
    x++;
    y++;
  }
  return z;
}
const long long int maxm = 1e6 + 3;
const long long int mod = 1e9 + 7;
const long long int inf = 1e18;
vector<long long int> fact, cnt;
void primeFactors(long long int n) {
  for (long long int i = 2; i < sqrt(n) + 1; i++) {
    if (n % i == 0) {
      fact.push_back(i);
      cnt.push_back(0);
    }
    while (n % i == 0) {
      cnt[fact.size() - 1]++;
      n = n / i;
    }
  }
  if (n > 2) {
    fact.push_back(n);
    cnt.push_back(1);
  }
}
void solve() {
  string s;
  cin >> s;
  string x = "";
  vector<string> v;
  for (long long int i = s.size() - 1; i > -1; i--) {
    if (s[i] != '0') v.push_back(s[i] + x);
    x = x + '0';
  }
  reverse(v.begin(), v.end());
  cout << v.size() << endl;
  for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
signed main() {
  ritik();
  long long int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}