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 mo = 1e9 + 7;
  long long int a[100];
  long long fastPower(long long base, long long power) {
    long long result = 1;
    while (power > 0) {
      if (power & 1) {
        result = result * base;
      }
      power >>= 1;
      base = (base * base);
    }
    return result;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      unsigned long long int n, k, ans = 0;
      cin >> n >> k;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
      }
      bool flag = false;
      for (int i = 1; i <= n; i++) {
        if (i == 1) {
          if (a[i] != 0) {
            ans = 1;
            flag = true;
            break;
          }
        } else {
          if (k < fastPower(10ll, a[i] - a[i - 1]) - 1) {
            ans += (k + 1) * fastPower(10ll, a[i - 1]);
            flag = true;
            break;
          } else {
            for (int j = a[i - 1]; j < a[i]; j++) {
              ans += fastPower(10ll, j) * 9;
            }
            k = k - fastPower(10ll, a[i] - a[i - 1]) + 1;
          }
        }
      }
      if (flag) {
        cout << ans << "\n";
        ;
      } else {
        ans += (k + 1) * fastPower(10ll, a[n]);
        cout << ans << "\n";
        ;
      }
    }
    return 0;
  }
  