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;
  using ll = long long;
  template <typename T>
  void print(T val) {
    cout << val;
  }
  template <typename T1, typename... T>
  void print(T1 x, T... args) {
    cout << x << ' ';
    print(args...);
  }
  template <typename... T>
  void println(T... args) {
    print(args..., '\n');
  }
  template <typename T>
  vector<T> readVec(int size) {
    vector<T> v(size);
    for (int i = 0; i < size; i++) {
      cin >> v[i];
    }
    return v;
  }
  template <typename T>
  void printVec(vector<T> v) {
    for (auto it : v) {
      cout << it << ' ';
    }
    cout << '\n';
  }
  void pYesNo(bool x) {
    if (x)
      cout << "Yes\n";
    else
      cout << "No\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll p[18] = {0};
    p[0] = 1;
    for (int i = 1; i < 18; i++) {
      p[i] = p[i - 1] * 10;
    }
    int t;
    cin >> t;
    while (t--) {
      ll n, k;
      cin >> n >> k;
      k++;
      auto v = readVec<int>(n);
      v.push_back(18);
      n++;
      ll ans = 0;
      for (int i = 1; i < n; i++) {
        ll d = v[i] - v[i - 1];
        ll x = p[d] - 1;
        if (x <= k) {
          k -= x;
          ans += p[v[i - 1]] * x;
        } else {
          ans += p[v[i - 1]] * k;
          break;
        }
      }
      println(ans);
    }
    return 0;
  }
  