description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: #include <bits/stdc++.h>
  using namespace std;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> arr;
      for (long long i = 0; i < n; i++) {
        long long h;
        cin >> h;
        long long topush = 1;
        while (h--) topush *= 10;
        arr.push_back(topush);
      }
      vector<long long> limits;
      for (long long i = 0; i < n - 1; i++) {
        limits.push_back(arr[i + 1] / arr[i] - 1);
      }
      long long rem = k + 1;
      long long current = 0;
      for (long long i = 0; i < n - 1; i++) {
        if (rem <= limits[i]) {
          current += arr[i] * rem;
          rem = 0;
          break;
        }
        current += (arr[i] * limits[i]);
        rem -= limits[i];
      }
      cout << current + (rem * arr[n - 1]) << "\n";
    }
    return 0;
  }
  