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;
  template <class T, template <class T2, class A = allocator<T2> > class cont>
  inline ostream &operator<<(ostream &out, const cont<T> &x) {
    for (const auto &it : x) {
      out << it << " ";
    }
    return out;
  }
  template <class T, template <class T2, class A = allocator<T2> > class cont>
  inline istream &operator>>(istream &in, cont<T> &x) {
    for (auto &it : x) {
      in >> it;
    }
    return in;
  }
  template <class T, class T2>
  inline ostream &operator<<(ostream &out, const pair<T, T2> &x) {
    out << x.first << " " << x.second;
    return out;
  }
  template <class T, class T2>
  inline istream &operator>>(istream &in, pair<T, T2> &x) {
    in >> x.first >> x.second;
    return in;
  }
  template <class T, class T2>
  inline bool chkmax(T &x, const T2 &y) {
    return x < y ? x = y, 1 : 0;
  }
  template <class T, class T2>
  inline bool chkmin(T &x, const T2 &y) {
    return x > y ? x = y, 1 : 0;
  }
  const long long mod = 1e9 + 7;
  const long long MAX_N = 100;
  long long arr[MAX_N];
  long long n, k;
  void solve() {
    cin >> n >> k;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
      long long pw = 1;
      while (arr[i]) {
        pw = (pw * 10ll);
        arr[i]--;
      }
      arr[i] = pw;
    }
    sort(arr, arr + n);
    k++;
    long long ret = 0;
    for (long long i = 0; i < n - 1; i++) {
      long long can = arr[i + 1] / arr[i] - 1;
      can = min(can, k);
      k -= can;
      ret += can * arr[i];
    }
    ret += k * arr[n - 1];
    cout << ret << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  