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;
  int a[12];
  long long b[12];
  long long pre[12];
  long long mul[21];
  void init() {
    long long x = 1;
    for (int i = 0; i <= 20; i++) {
      mul[i] = x;
      x *= 10;
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[i] = mul[a[i]];
    }
    long long dps = k + 1, pos = 1;
    long long ans = 0;
    while (dps > 0 and pos <= n) {
      long long tmp = pos < n ? min(dps, b[pos + 1] / b[pos] - 1) : dps;
      ans += b[pos] * tmp;
      pos++;
      dps -= tmp;
    }
    cout << ans << endl;
  }
  int main() {
    init();
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  