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;
  bool conv(long long n, const vector<long long>& a, long long k) {
    long long counter = 0;
    for (int i = a.size() - 1; i >= 0; --i) {
      counter += (n / a[i]);
      n -= (n / a[i]) * a[i];
      if (n <= 0) {
        break;
      }
    }
    if (counter > k) {
      return true;
    }
    return false;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      a[i] = (long long)pow(10, a[i]);
    }
    long long res = 0;
    long long counter = 0;
    for (int i = 0; i < n - 1; ++i) {
      res += ((a[i + 1] / a[i]) - 1) * a[i];
      counter += ((a[i + 1] / a[i]) - 1);
      if (counter > k) {
        while (counter > k) {
          res -= a[i];
          counter--;
        }
        res += a[i];
        counter++;
      }
    }
    if (counter <= k) {
      res += (k + 1 - counter) * a[n - 1];
    }
    cout << res << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  