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;
  const int N = 10;
  int aa[N];
  long long pp[10];
  int main() {
    long long p = 1;
    for (int h = 0; h < 10; h++, p *= 10) pp[h] = p;
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k, k++;
      for (int i = 0; i < n; i++) cin >> aa[i];
      long long x = 0;
      for (int i = 1; i < n; i++) {
        int l = aa[i - 1], r = aa[i];
        if (k >= pp[r - l] - 1) {
          x += r - l;
          k -= pp[r - l] - 1;
        } else
          break;
      }
      if (k > 0) cout << k;
      while (x--) cout << 9;
      cout << '\n';
    }
    return 0;
  }
  