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 ckmin(T &a, const T &b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  void re(T &x) {
    cin >> x;
  }
  template <class Arg, class... Args>
  void re(Arg &first, Args &...rest) {
    re(first);
    re(rest...);
  }
  template <class T>
  void re(vector<T> &a) {
    for (int i = 0; i < (int)(a.size()); ++i) re(a[i]);
  }
  template <class T>
  void pr(T &x) {
    cout << x << " ";
  }
  template <class Arg, class... Args>
  void pr(Arg &first, Args &...rest) {
    pr(first);
    pr(rest...);
  }
  template <class T>
  void pr(vector<T> &a) {
    for (int i = 0; i < (int)(a.size()); ++i) pr(a[i]);
  }
  void solve() {
    unsigned long long n, k;
    cin >> n >> k;
    vector<long long> a(10);
    for (int i = 0; i < n; i++) {
      int pos;
      cin >> pos;
      a[pos] = 1;
    }
    vector<long long> mult(18);
    unsigned long long curmult = 1;
    for (int i = 0; i < 18; i++) {
      mult[i] = curmult * 9;
      if (i + 1 < 10 && a[i + 1] == 1)
        curmult = 1;
      else
        curmult *= 10;
    }
    vector<long long> ans(18);
    for (int i = 17; i >= 0; i--) {
      unsigned long long rest = 0;
      for (int j = 0; j < i; j++) rest += mult[j];
      for (int j = i + 1; j < 18; j++) rest += mult[j];
      unsigned long long d = 9;
      while (d > 0 && rest + (mult[i] / 9) * (d - 1) > k) {
        d--;
      }
      if (rest + (mult[i] / 9) * (d) <= k) d++;
      mult[i] = (mult[i] / 9) * d;
      ans[i] = d;
    }
    unsigned long long make = 1;
    unsigned long long made = 0;
    for (int i = 0; i < (int)(18); ++i) {
      made += ans[i] * make;
      make *= 10;
    }
    pr(made);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(10);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  