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() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> v(n);
      for (long long i = 0; i < n; i++) {
        long long x;
        cin >> x;
        v[i] = pow(10LL, x);
      }
      sort(v.begin(), v.end());
      vector<long long> diff;
      for (long long i = 1; i < n; i++) {
        diff.push_back(v[i] / v[i - 1] - 1);
      }
      long long sum = 0, idx = -1, sum1 = 0;
      for (long long i = 0; i < (long long)diff.size(); i++) {
        sum += diff[i];
        if (sum > k) {
          sum1 = sum - diff[i];
          idx = i;
          break;
        }
      }
      if (idx == -1) {
        long long rem = (k - sum + 1) * (*prev(v.end()));
        if ((long long)v.size() == 1) {
          cout << rem << "\n";
        } else {
          cout << rem + (*prev(v.end())) - 1 << "\n";
        }
      } else {
        long long rem = (k - sum1 + 1) * v[idx];
        cout << rem + v[idx] - 1 << "\n";
      }
    }
  }
  