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;
  template <class T>
  bool uin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
  const int maxN = 1e5 + 10;
  int n, a[maxN];
  long long k, pw[19];
  void solve() {
    cin >> n >> k;
    for (int i = 0; i < (int)(n); ++i) {
      cin >> a[i];
    }
    pw[0] = 1;
    for (int i = 1; i <= (int)(18); ++i) {
      pw[i] = pw[i - 1] * 10;
    }
    k++;
    string ans;
    for (int i = 1; i <= (int)(n - 1); ++i) {
      int d = a[i] - a[i - 1];
      if (pw[d] - 1 < k) {
        ans += string(d, '9');
        k -= pw[d] - 1;
      } else {
        string t = to_string(k);
        reverse((t).begin(), (t).end());
        ans += t;
        k = 0;
        break;
      }
    }
    if (k) {
      string t = to_string(k);
      reverse((t).begin(), (t).end());
      ans += t;
    }
    reverse((ans).begin(), (ans).end());
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  