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 long long maxn = 1e2 + 10;
  long long a[maxn], bin[maxn], can[maxn];
  long long n, k, ans;
  signed main() {
    bin[0] = 1;
    for (register long long i = 1; i <= 9; ++i) bin[i] = bin[i - 1] * 10;
    long long T;
    cin >> T;
    while (T--) {
      cin >> n >> k;
      for (register long long i = 1; i <= n; ++i) cin >> a[i];
      for (register long long i = 1; i < n; ++i)
        can[i] = (bin[a[i + 1]] - bin[a[i]]) / bin[a[i]];
      ans = 0;
      for (register long long i = 1; i < n; ++i) {
        if (can[i] <= k) {
          k -= can[i];
          ans += can[i] * bin[a[i]];
          if (!k) ans += bin[a[i + 1]];
        } else
          ans += (k + 1) * bin[a[i]], k = 0;
        if (!k) break;
      }
      if (k) ans += bin[a[n]] * (k + 1);
      cout << ans;
      puts("");
    }
    return 0;
  }
  