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;
  void solve() {
    long long n, k;
    cin >> n >> k;
    long long a[n], atmax[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long notes = 0, sum = 0;
    for (long long i = 0; i < n - 1; i++) {
      long long v1 = pow(10, a[i]);
      long long v2 = pow(10, a[i + 1]);
      long long val = v2 / v1 - 1;
      atmax[i] = val;
    }
    for (long long i = 0; i < n; i++) {
      if (i == n - 1) {
        long long left = k - notes + 1;
        long long val = pow(10, a[i]);
        sum += (val * left);
        break;
      }
      if (notes + atmax[i] <= k) {
        long long val = pow(10, a[i]);
        sum += (val * atmax[i]);
        notes += atmax[i];
      } else {
        long long left = k - notes + 1;
        long long val = pow(10, a[i]);
        sum += (val * left);
        break;
      }
    }
    cout << sum << '\n';
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) solve();
  }
  