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 inf = 1000000007;
  long long int facto[1000010];
  long long int power(long long int a, long long int pow) {
    long long int ans = 1;
    if (a < 0) a += inf;
    while (pow) {
      if (pow % 2 == 1) {
        ans = ((ans % inf) * (a % inf)) % inf;
      }
      a = ((a % inf) * (a % inf)) % inf;
      if (a < 0) a += inf;
      pow /= 2;
    }
    return ans;
  }
  void fact(int n) {
    long long int fac = 1;
    for (int i = 1; i <= n; i++) {
      fac = ((fac % inf) * (i % inf)) % inf;
      facto[i] = fac;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    facto[0] = 1;
    int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      int arr[n];
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      k += 1;
      long long int val = 0;
      for (int i = 0; i < n - 1; i++) {
        long long int x = power(10, arr[i + 1] - arr[i]) - 1;
        if (k > x) {
          val += x * power(10, arr[i]);
          k -= x;
        } else {
          val += k * power(10, arr[i]);
          k = 0;
          break;
        }
      }
      if (k > 0) val += k * power(10, arr[n - 1]);
      cout << val << "\n";
    }
    return 0;
  }
  