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;
  using Vi = vector<int>;
  using i64 = long long;
  using ll = long long;
  using Pi = pair<int, int>;
  mt19937 mrand(time(0));
  int rnd(int x) { return mrand() % x; }
  const int N = 500005;
  const int M = 40005;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n;
      ll k;
      cin >> n >> k;
      ++k;
      Vi a(n);
      vector<ll> ten(15);
      ten[0] = 1;
      for (int i = 1; i < 15; ++i) ten[i] = ten[i - 1] * 10;
      for (int i = 0; i < n; ++i) cin >> a[i];
      ll ans = 0;
      for (int i = 0; i < n; ++i) {
        ll t = (i == n - 1 ? 1000000001 : ten[a[i + 1] - a[i]] - 1);
        ll tmd = min(t, k);
        ans += tmd * ten[a[i]];
        k -= tmd;
      }
      cout << ans << endl;
    }
    return 0;
  }
  