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;
  long long int power(long long int a, long long int b) {
    long long int result;
    if (b == 0) return 1;
    if (b % 2 != 0) {
      result = a;
      result = (result * (power(a, b - 1)));
    } else {
      result = power(a, b / 2);
      result = (result * result);
    }
    return (long long int)result;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      long long int sum = 0;
      long long int prev = 0;
      vector<long long int> v(n);
      for (auto &i : v) cin >> i;
      bool f = 1;
      for (int i = 0; i < n - 1; i++) {
        long long int temp = power(10, v[i + 1] - v[i]) - 1;
        if (temp <= k) {
          sum += temp * (power(10, v[i]));
          k -= temp;
        } else {
          sum += (k + 1) * (power(10, v[i]));
          f = 0;
          break;
        }
      }
      if (f) {
        sum += (k + 1) * (power(10, v[n - 1]));
      }
      cout << sum << "\n";
    }
    return 0;
  }
  