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 fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  int main() {
    fastio();
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long a[n];
      long long i, j;
      set<long long> st;
      for (i = 0; i < n; i++) {
        cin >> a[i];
        st.insert(a[i]);
      }
      long long val[18] = {0};
      for (i = 0; i < 18; i++) {
        if (st.find(i) != st.end()) {
          val[i] = 1;
        }
      }
      long long cur = 10;
      for (i = 1; i < 18; i++) {
        if (val[i] == 0) {
          val[i] = cur;
          cur = cur * 10;
        } else {
          cur = 10;
        }
      }
      long long cnt = 0;
      cur = 0;
      vector<long long> pp;
      pp.push_back(0);
      while (true) {
        long long p = val[cnt] * 9;
        cur += p;
        cnt++;
        pp.push_back(cur);
        if (cur > k) {
          break;
        }
      }
      string ans = "";
      long long tot = cur;
      for (i = 0; i < cnt; i++) {
        cur = pp.back();
        pp.pop_back();
        long long v = pp.back();
        for (j = 0; j <= 9; j++) {
          if (val[cnt - i - 1] * j + v > k) {
            break;
          }
        }
        ans += '0' + j;
        k -= val[cnt - i - 1] * j;
      }
      cout << ans << "\n";
    }
  }
  