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;
  long long n, k;
  long long a[15];
  long long dp[15];
  long long ans[15];
  string an;
  signed main() {
    ios::sync_with_stdio(0);
    long long T;
    cin >> T;
    while (T--) {
      memset(dp, 0, sizeof dp);
      memset(ans, 0, sizeof ans);
      memset(a, 0, sizeof a);
      cin >> n >> k;
      for (long long i = 1; i <= n; i++) {
        cin >> a[i];
      }
      for (long long i = 1; i <= n - 1; i++) {
        dp[i] = pow(10ll, a[i + 1] - a[i]) - 1ll;
      }
      long long so = k + 1;
      for (long long i = 1; i <= n - 1; i++) {
        if (so >= dp[i]) {
          so -= dp[i];
          ans[i] = dp[i];
        } else {
          ans[i] = so;
          so = 0;
          break;
        }
      }
      an = "";
      for (long long i = 1; i < n; i++) {
        if (ans[i] == 0) break;
        string t = to_string(ans[i]);
        reverse(t.begin(), t.end());
        an += t;
      }
      if (so > 0) {
        string t = to_string(so);
        reverse(t.begin(), t.end());
        an += t;
      }
      reverse(an.begin(), an.end());
      cout << an << endl;
    }
  }
  