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;
  void solve() {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> v(n);
    for (int i = 0; i < n; i++) cin >> v[i];
    ;
    sort(v.begin(), v.end());
    vector<long long int> p(10, 0);
    p[0] = 1;
    for (int i = 1; i <= 9; i++) {
      p[i] = p[i - 1] * 10;
    }
    long long int s = 2e18;
    long long int ans = 0;
    long long int cnt = 0;
    k++;
    for (int i = 0; i < n; i++) {
      if (i == n - 1 || ans + k * p[v[i]] < p[v[i + 1]]) {
        long long int val = ans + k * p[v[i]];
        s = min(val, s);
      }
      long long int diff = 0;
      if (i != n - 1) {
        diff = v[i + 1] - v[i];
      }
      ans += min(p[diff] - 1, k) * p[v[i]];
      k -= min(p[diff] - 1, k);
    }
    cout << s << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  