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 double pi = acos(-1);
  long long q_pow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  const int N = 2e5 + 456;
  long long a[100], dp[12], b[100];
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        if (!x) {
          a[i] = 1;
        } else {
          long long sum = 1;
          for (int j = 1; j <= x; j++) {
            sum *= 10;
          }
          a[i] = sum;
        }
      }
      a[n + 1] = 2 * a[n];
      long long ans = 0, sum = 0;
      for (int i = 1; i <= n; i++) {
        b[i] = (a[i + 1] / a[i] - 1);
      }
      int ok = 0;
      for (int i = 1; i <= n - 1; i++) {
        if (k - ans < b[i]) {
          ok = 1;
          sum += (k - ans + 1) * a[i];
          break;
        } else {
          ans += b[i];
          sum += b[i] * a[i];
        }
      }
      if (ok)
        cout << sum << '\n';
      else {
        sum += (k - ans + 1) * a[n];
        cout << sum << '\n';
      }
    }
    return 0;
  }
  