description: On the board, Bob wrote n positive integers in [base](https://en.wikipedia.org/wiki/Positional_notation#Base_of_the_numeral_system) 10 with sum s (i. e. in decimal numeral system). Alice sees the board, but accidentally interprets the numbers on the board as base-11 integers and adds them up (in base 11).
  
  What numbers should Bob write on the board, so Alice's sum is as large as possible?
  
  Input
  
  The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
  
  The only line of each test case contains two integers s and n (1 ≤ s ≤ 10^9; 1 ≤ n ≤ min(100, s)) — the sum and amount of numbers on the board, respectively. Numbers s and n are given in decimal notation (base 10).
  
  Output
  
  For each test case, output n positive integers — the numbers Bob should write on the board, so Alice's sum is as large as possible. If there are multiple answers, print any of them.
  
  Example
  
  Input
  
  
  6
  97 2
  17 1
  111 4
  100 2
  10 9
  999999 3
  
  
  Output
  
  
  70 27 
  17 
  3 4 100 4
  10 90
  1 1 2 1 1 1 1 1 1 
  999900 90 9
  
  Note
  
  In the first test case, 70_{10} + 27_{10} = 97_{10}, and Alice's sum is $$$70_{11} + 27_{11} = 97_{11} = 9 ⋅ 11 + 7 = 106_{10}. (Here x_b represents the number x in base b.) It can be shown that it is impossible for Alice to get a larger sum than 106_{10}$$$.
  
  In the second test case, Bob can only write a single number on the board, so he must write 17.
  
  In the third test case, 3_{10} + 4_{10} + 100_{10} + 4_{10} = 111_{10}, and Alice's sum is $$$3_{11} + 4_{11} + 100_{11} + 4_{11} = 110_{11} = 1 ⋅ 11^2 + 1 ⋅ 11 = 132_{10}. It can be shown that it is impossible for Alice to get a larger sum than 132_{10}$$$.
solution: #include <bits/stdc++.h>
  using namespace std;
  int xm[4] = {-1, 1, 0, 0};
  int ym[4] = {0, 0, -1, 1};
  const int MOD = 1e9 + 7;
  const int MAXN = 5e5 + 5;
  const long long POW = 9973;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int flg = 0;
    int t;
    cin >> t;
    while (t--) {
      long long second, n;
      cin >> second >> n;
      if (second == 202) flg = 1;
      long long temp = second;
      long long rem = second;
      vector<long long> v;
      while (second) {
        v.push_back(second % 10);
        second /= 10;
      }
      vector<long long> ans;
      long long p10 = 1;
      for (long long i = 1; i < v.size(); i++) p10 *= 10;
      for (long long i = v.size() - 1; i >= 0; i--) {
        for (long long j = 10 * n; j >= 0; j--) {
          if (rem - j * p10 >= 0 && j + (rem - (j * p10)) >= n - ans.size()) {
            vector<int> cnt(n, 0);
            rem -= j * p10;
            int q = j;
            while (ans.size() < n) {
              if (q) {
                ans.push_back(p10);
                q--;
                cnt[ans.size() - 1]++;
              } else
                break;
            }
            for (int i = 0; i < ans.size(); i++) {
              if (q) {
                int ad = min(9 - cnt[i], q);
                ans[i] += ad * p10;
                q -= ad;
              }
            }
            break;
          }
        }
        p10 /= 10;
      }
      if (rem) cout << temp << " " << n << "\n";
      for (auto i : ans) cout << i << " ";
      ;
      cout << "\n";
    }
  }
  