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 int mod = 1e9 + 7;
  void yes() { cout << "YES" << endl; }
  void no() { cout << "NO" << endl; }
  bool compare(pair<long long int, pair<long long int, long long int> > &a,
               pair<long long int, pair<long long int, long long int> > &b) {
    if (a.first == b.first) {
      return a.second.second <= b.second.second;
    }
    return a.first < b.first;
  }
  bool isPrime(long long int n) {
    for (int i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        return 0;
      }
    }
    return 1;
  }
  long long int solve() {
    long long int n, k, m;
    cin >> n >> k;
    string s, t = "", r;
    long long int count = 0, prev = 0, cnt = 0, ans = 0, curr = 0;
    bool turn = 1, win = 0, flag = 1;
    vector<long long int> v(n, 0);
    for (int i = 0; i < n; i++) cin >> v[i];
    if (v[0] != 0) {
      ans = 1;
    } else if (n == 1) {
      ans = k + 1;
    } else {
      count = k;
      for (int i = 1; i < n; i++) {
        cnt = (long long int)pow((long long int)10, v[i]) -
              (long long int)pow((long long int)10, v[i - 1]);
        if (count * (long long int)pow((long long int)10, v[i - 1]) < cnt) {
          ans += (count + 1) * (long long int)pow((long long int)10, v[i - 1]);
          count = -1;
          break;
        } else {
          ans += cnt;
          count -= cnt / (long long int)pow((long long int)10, v[i - 1]);
        }
      }
      ans += (count + 1) * (long long int)pow((long long int)10, v[n - 1]);
    }
    cout << ans << endl;
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    cin >> test;
    for (int i = 0; i < test; i++) {
      solve();
    }
  }
  