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;
  const int N = 20;
  int T, n, k;
  long long inf = 1e18;
  vector<int> v;
  void _clear() { v.clear(); }
  long long po(long long x, long long os) {
    if (os == 0) return 1;
    long long z = po(x, os / 2);
    if (os & 1) return z * z * x;
    return z * z;
  }
  long long calc(int k) {
    long long ret = 0;
    long long num = 0;
    for (int i = 0; i <= n - 2; i++) {
      long long moves = min(1LL * v[i + 1] / v[i] - 1, 1LL * k - num);
      ret += moves * v[i];
      num += moves;
    }
    ret += (k - num) * v.back();
    return ret;
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      _clear();
      scanf("%d%d", &n, &k);
      for (int i = 1; i <= n; i++) {
        int x;
        scanf("%d", &x);
        v.push_back(po(10, x));
      }
      sort(v.begin(), v.end());
      printf("%lld\n", calc(k + 1));
    }
  }
  