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 mod = 1e9 + 7;
  const long long inf = 1e16;
  const long long N = 2e5 + 10;
  long long pw(long long a, long long n) {
    if (n == 0) return 1;
    long long res = pw(a, n / 2);
    if (n % 2)
      return res * res * a;
    else
      return res * res;
  }
  void run() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (auto &e : a) {
      cin >> e;
    }
    reverse((a).begin(), (a).end());
    long long rem = 19;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      rem -= a[i];
      v.push_back(rem);
      rem = a[i];
    }
    reverse((a).begin(), (a).end());
    reverse((v).begin(), (v).end());
    long long ans = 0;
    k++;
    for (long long i = 0; i < n - 1; i++) {
      long long cur = stoll(string(v[i], '9'));
      if (cur > k) {
        ans += k * pw(10, a[i]);
        k = 0;
        break;
      } else {
        ans += cur * pw(10, a[i]);
        k -= cur;
      }
      if (k <= 0) break;
    }
    ans += k * 1ll * pw(10, a[n - 1]);
    cout << ans << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long tt = 1;
    cin >> tt;
    while (tt--) run();
    return 0;
  }
  