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 long long M = 1e9 + 7;
  const long long IM = 1e18 + 37;
  const long long N = 2e5 + 10;
  long long solve(long long cn, long long a[], long long n, long long k) {
    long long ans = 0;
    for (long long i = n - 1; i >= 0; i--) {
      long long val = 1, now = a[i];
      while (now--) val = val * 10;
      ans += cn / val;
      cn = cn - ((cn / val) * val);
    }
    return (ans > k);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      long long a[n];
      for (long long &it : a) cin >> it;
      long long ans = 0;
      long long val = 1;
      long long u = k;
      for (long long i = 0; i < n; i++) {
        long long id = 99999999999;
        if (i + 1 <= n - 1) {
          id = 0;
          long long cn = a[i + 1] - a[i];
          while (cn--) {
            id = id * 10 + 9;
          }
        }
        val = 1;
        long long now = a[i];
        while (now--) {
          val = val * 10;
        }
        if (solve(ans + (k + 1) * val, a, n, u)) {
          ans += (k + 1) * val;
          break;
        }
        id = min(k, id);
        k -= id;
        ans += id * val;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  