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;
  template <class T>
  inline T MAX(T a, T b) {
    if (a > b) return a;
    return b;
  }
  template <class T>
  inline T MIN(T a, T b) {
    if (a < b) return a;
    return b;
  }
  template <class T>
  inline T ABS(T x) {
    if (x < 0) return -x;
    return x;
  }
  inline void OPEN(const string &s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
  const static int INF = 2123123123;
  int tc;
  long long n, k;
  long long a[12];
  long long powI(long long a, long long b) {
    long long result = 1;
    for (int(i) = (0); (i) < (b); ++(i)) {
      result *= a;
    }
    return result;
  }
  int main() {
    scanf("%d", &tc);
    while (tc--) {
      scanf("%lld%lld", &n, &k);
      for (int(i) = (0); (i) < (n); ++(i)) {
        scanf("%lld", &a[i]);
      }
      long long result = 0;
      long long sisaK = k;
      for (int i = 0; i < n; i++) {
        if (i + 1 < n) {
          long long maxTime = powI(10, a[i + 1] - a[i]) - 1;
          long long times = min(maxTime, sisaK);
          sisaK -= times;
          result += powI(10, a[i]) * times;
          if (sisaK == 0 && powI(10, a[i]) * (times + 1) != powI(10, a[i + 1])) {
            result += powI(10, a[i]);
            break;
          }
        } else {
          result += powI(10, a[i]) * (sisaK + 1);
        }
      }
      printf("%lld\n", result);
    }
    return 0;
  }
  