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 setIO() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  const int64_t N = 1e5 + 5;
  const int64_t M = 650;
  const int64_t inf = 9e12;
  const int64_t mod = 998244353;
  int64_t xyz = 1;
  int64_t n, m;
  int64_t arr[N];
  int64_t pwr[N];
  void run() {
    cin >> n >> m;
    m++;
    pwr[0] = 1;
    for (int64_t i = 1; i <= 12; i++) {
      pwr[i] = pwr[i - 1] * 10;
    }
    for (int64_t i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    arr[n + 1] = 12;
    int64_t num = 0;
    int64_t sum = 0;
    for (int64_t i = 1; i <= n; i++) {
      int64_t add = pwr[arr[i + 1] - arr[i]] - 1;
      if (i == n || sum + add > m) {
        num += (m - sum) * pwr[arr[i]];
        break;
      } else
        sum += add, num += pwr[arr[i]] * add;
    }
    cout << num << "\n";
  }
  signed main() {
    setIO();
    cin >> xyz;
    while (xyz--) run();
    return 0;
  }
  