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 Fast() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  void File() {}
  using ll = long long;
  using ull = unsigned long long;
  int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
  int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
  const int N = 1e6 + 5;
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    for (auto& a : v) cin >> a;
    string ans;
    int bit = 0;
    ll num = 9;
    k++;
    for (int i = 0; k && i < n - 1; i++, num = 9, bit++) {
      while (v[i + 1] != bit + 1) {
        num = num * 10 + 9;
        bit++;
      }
      if (i == n) break;
      if (k <= num) {
        ans = to_string(k) + ans;
        k = 0;
        break;
      } else {
        ans = to_string(num) + ans;
        k -= num;
      }
    }
    if (k) cout << k;
    cout << ans << "\n";
  }
  int main() {
    Fast();
    File();
    int t$$t$;
    cin >> t$$t$;
    while (t$$t$--) solve();
  }
  