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 min(long long a, long long b) { return a < b ? a : b; }
  long long max(long long a, long long b) { return a > b ? a : b; }
  bool isprime(long long k) {
    bool is = 1;
    for (long long i = 2; i * i <= k; i++)
      if (k % i == 0) is = 0;
    return k > 1 ? is : 0;
  }
  const double PI = acos(-1);
  long long n, k;
  long long a[15];
  inline void sol() {
    cin >> n >> k;
    k++;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    if (n == 1) {
      cout << k << "\n";
      return;
    }
    long long tot = 0;
    for (long long i = 2; i <= n; i++) {
      long long pw = 1;
      for (long long j = a[i - 1]; j < a[i]; j++) pw *= 10;
      pw -= 1;
      if (tot + pw >= k) {
        long long res = 0;
        pw = 1;
        for (long long j = 0; j < a[i - 1]; j++) res = res * 10 + 9, pw *= 10;
        k -= tot;
        cout << k * pw + res << "\n";
        return;
      } else
        tot += pw;
    }
    long long res = 0, pw = 1;
    for (long long i = 0; i < a[n]; i++) res = res * 10 + 9, pw *= 10;
    cout << (k - tot) * pw + res << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long _ = 1;
    cin >> _;
    while (_--) sol();
    return 0;
  }
  