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 int power(long long int a, long long int b, long long int mod) {
    long long int ans = 1;
    while (b > 0) {
      if (b & 1) {
        ans = (ans * a) % mod;
      }
      a = (a * a) % mod;
      b = b / 2;
    }
    return ans;
  }
  int main() {
    vector<long long int> temp(11);
    temp[0] = 1;
    for (long long int i = 1; i <= 10; i++) temp[i] = temp[i - 1] * 10;
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
      long long int n, k;
      cin >> n >> k;
      vector<long long int> arr(n);
      k++;
      for (long long int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      long long int check = n - 1;
      long long int val = 0;
      for (long long int i = 0; i < n - 1; i++) {
        val += (temp[arr[i + 1] - arr[i]] - 1);
        if (val >= k) {
          check = i;
          break;
        }
      }
      long long int ans = 0;
      for (long long int j = 0; j <= check; j++) {
        if (j == check) {
          ans += (temp[arr[j]] * k);
          break;
        }
        long long int ta = (temp[arr[j + 1] - arr[j]] - 1);
        if (ta >= k) {
          ans += (temp[arr[j]] * k);
          break;
        } else {
          ans += (temp[arr[j]] * ta);
          k = k - ta;
        }
      }
      cout << ans << endl;
    }
  }
  