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 MAX = 1e5 + 10;
  const long long LMAX = 1e9 + 10;
  const long double LDMAX = 1e9 + 10;
  const long long MOD = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  const long long LLINF = 0x3f3f3f3f3f3f3f3f;
  const long double PI = acos(-1);
  int32_t main(void) {
    ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    ;
    long long n, m, t, k;
    long long a, i, j;
    long long ans = 0;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      vector<long long> a(n);
      for (i = 0; i < n; i++) {
        cin >> a[i];
      }
      long long last = 0;
      k += 1;
      ans = 0;
      for (i = 0; i < n; i++) {
        long long temp = (a[i] - last);
        if (temp == 0) continue;
        long long num = 0;
        m = 1;
        for (j = 0; j < temp; j++) {
          num += 9 * m;
          m *= 10;
        }
        if (k >= num) {
          k -= num;
          for (j = 1; j <= a[i - 1]; j++) num *= 10;
          ans += num;
        } else {
          num = k;
          for (j = 1; j <= a[i - 1]; j++) num *= 10;
          ans += num;
          k = 0;
          break;
        }
        last = a[i];
      }
      if (k > 0) {
        long long num = k;
        for (j = 1; j <= a[n - 1]; j++) num *= 10;
        ans += num;
      }
      cout << ans << '\n';
    }
  }
  