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 _out() { cerr << '\n'; }
  template <typename H, typename... T>
  void _out(H h, T... t) {
    cerr << ' ' << h;
    _out(t...);
  }
  void solve();
  void init();
  signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    long long T = 1;
    cin >> T;
    init();
    while (T--) {
      solve();
    }
    return 0;
  }
  long long const N = 20;
  long long a[N];
  long long ten[N];
  void init() {
    long long t = 1;
    for (long long i = 0; i < N; i++) {
      ten[i] = t;
      t *= 10;
    }
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long ans = 0;
    k++;
    for (long long i = 0; i < n; i++) {
      if (i != n - 1 && k >= ten[a[i + 1] - a[i]] - 1) {
        ans += (ten[a[i + 1] - a[i]] - 1) * ten[a[i]];
        k -= ten[a[i + 1] - a[i]] - 1;
      } else {
        ans += k * ten[a[i]];
        break;
      }
    }
    cout << ans << '\n';
  }
  