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 T, k, n, a[24], las, now, sum;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T--) {
      cin >> k >> n;
      n++;
      memset(a, 0, sizeof(a));
      for (long long i = 1; i <= k; ++i) {
        long long f;
        cin >> f;
        a[f] = true;
      }
      if (n <= 9) {
        cout << n << endl;
        continue;
      }
      las = 0;
      now = 9;
      sum = 0;
      for (long long i = 1; i <= 19; ++i)
        if (a[i] == 0) {
          now = now * 10 + 9;
          if (now + sum >= n) {
            cout << n - sum;
            for (long long j = las - 1; j >= 0; j--) cout << '9';
            break;
          }
        } else {
          las = i;
          sum += now;
          now = 9;
          if (sum + now >= n) {
            cout << n - sum;
            for (long long j = i - 1; j >= 0; j--) cout << 9;
            break;
          }
        }
      cout << endl;
    }
  }
  