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;
  long long T, n, k;
  long long a[100], s[100], pre[18];
  signed main() {
    scanf("%lld", &T);
    pre[0] = 1;
    for (long long i = 1; i <= 18; i++) pre[i] = pre[i - 1] * 10;
    while (T--) {
      scanf("%lld%lld", &n, &k);
      k++;
      for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]), s[i] = 0;
      for (long long i = 1; i <= n - 1; i++) {
        long long t = pre[a[i + 1] - a[i]] - 1;
        if (t < k) {
          s[i] = t;
          k -= t;
        } else {
          s[i] = k;
          k = 0;
          break;
        }
      }
      if (k) s[n] = k;
      long long ans = 0;
      for (long long i = 1; i <= n; i++) ans += s[i] * pre[a[i]];
      printf("%lld\n", ans);
    }
  }
  