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(false);
    cin.tie(0);
    array<__int128, 21> p10 = {};
    p10[0] = 1;
    for (int i = 1; i < 21; i++) {
      p10[i] = p10[i - 1] * 10;
    }
    long long tt;
    cin >> tt;
    while (tt--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> a(n);
      array<long long, 21> mp = {};
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
        mp[a[i]] = 1;
      }
      auto f = [&](__int128 x) {
        __int128 ans = 0;
        for (int i = 20; i >= 0; i--) {
          __int128 d = x / p10[i];
          x -= d * p10[i];
          if (!d) continue;
          __int128 c = 0;
          for (int j = i; j >= 0; j--) {
            if (mp[j]) {
              ans += d * p10[c];
              break;
            }
            c += 1;
          }
        }
        return ans;
      };
      __int128 nn = 0;
      vector<__int128> v(21);
      for (long long j = 0; j < 21; j++) {
        nn = 10LL * nn + 9LL;
      }
      if (f(nn) < k) {
        assert(false);
      }
      long long cj = 20;
      while (f(nn) > k && cj >= 0) {
        while (nn - p10[cj] > 0 && f(nn - p10[cj]) > k) {
          nn -= p10[cj];
        }
        cj -= 1;
      }
      cout << (long long)nn << '\n';
    }
    return 0;
  }
  