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 a[5007], b[5007];
  long long pre[5007], back[5007];
  long long ksm(long long base, long long p) {
    long long r = 1;
    while (p > 0) {
      if (p & 1) {
        r = r * base;
      }
      p >>= 1;
      base = (base * base);
    }
    return r;
  }
  int main() {
    ios::sync_with_stdio(false);
    long long T;
    cin >> T;
    while (T--) {
      long long n, m;
      cin >> n >> m;
      for (long long i = 1; i <= n; i++) {
        long long x;
        cin >> x;
        a[i] = ksm(10, x);
        b[i] = x;
      }
      long long ans = 0;
      long long t = m + 1;
      for (long long i = 1; i < n; i++) {
        long long z = b[i + 1] - b[i];
        long long q = ksm(10, z) - 1;
        if (t - q >= 0) {
          t = t - q;
          ans += q * a[i];
        } else {
          ans += t * a[i];
          t = 0;
          break;
        }
      }
      if (t != 0) ans += t * a[n];
      cout << ans << endl;
    }
    return 0;
  }
  