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;
  signed main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> a(n);
      for (long long i = 0; i < n; i++) cin >> a[i];
      if (n == 1) {
        cout << k + 1 << "\n";
        continue;
      }
      k++;
      bool ok = false;
      long long new_delta = 0, m = 0;
      for (m = 1; m < n && !ok; m++) {
        new_delta = pow(10, a[m] - a[m - 1]) - 1;
        if (new_delta < k)
          k -= new_delta;
        else
          ok = true;
      }
      stringstream ss;
      if (k) ss << k;
      if (m >= 2) {
        if (ok)
          m = a[m - 2];
        else
          m = a[n - 1];
        for (long long i = 0; i < m; i++) ss << "9";
      }
      cout << ss.str() << "\n";
    }
    return 0;
  }
  