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 = 11;
  long long a[maxn];
  long long b[maxn];
  inline long long qpow(long long x, long long y) {
    long long res(1);
    while (y) {
      if (y & 1) res = 1ll * res * x;
      x = 1ll * x * x;
      y >>= 1;
    }
    return res;
  }
  signed main() {
    ios::sync_with_stdio(false);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      k++;
      for (long long i = 1; i <= n; i++) cin >> a[i];
      for (long long i = 2; i <= n; i++) b[i] = a[i] - a[i - 1];
      long long now = 0;
      for (long long i = 2; i <= n; i++) {
        long long tmp = qpow(10, b[i]) - 1;
        if (k > tmp) {
          k -= tmp;
          now += tmp * qpow(10, a[i - 1]);
        } else {
          now += k * qpow(10, a[i - 1]);
          k = 0;
          break;
        }
      }
      if (k > 0) {
        now += k * qpow(10, a[n]);
      }
      cout << now << "\n";
    }
  }
  