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 a[15], b[15];
  long long lt(long long x, long long y) {
    if (y == 0) return 1;
    long long ans = 1;
    for (long long i = 1; i <= y; i++) {
      ans *= x;
    }
    return ans;
  }
  int main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, k, max = 0, ans = 0;
      cin >> n >> k;
      b[n] = 0;
      for (long long i = 1; i <= n; i++) {
        cin >> a[i];
        if (i >= 2) {
          b[i - 1] = lt(10, a[i] - a[i - 1]) - 1;
          max += b[i - 1];
        }
      }
      if (max > k + 1) {
        for (long long i = n - 1; i >= 1; i--) {
          long long sub = min(max - k - 1, b[i]);
          max -= sub;
          b[i] -= sub;
          if (max == k + 1) {
            break;
          }
        }
      } else if (max <= k) {
        b[n] = k - max + 1;
      }
      for (long long i = 1; i <= n; i++) {
        ans += lt(10, a[i]) * b[i];
      }
      cout << ans << endl;
    }
    return 0;
  }
  