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 gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
  bool sorta(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.second < b.second);
  }
  bool sortd(const pair<int, int> &a, const pair<int, int> &b) {
    return (a.second > b.second);
  }
  void printarr(long long arr[], long long n) {
    for (int i = 0; i < n; i++) cout << arr[i] << " ";
    cout << "\n";
  }
  string decToBinary(int n) {
    string s = "";
    int i = 0;
    while (n > 0) {
      s = to_string(n % 2) + s;
      n = n / 2;
      i++;
    }
    return s;
  }
  long long binaryToDecimal(string n) {
    string num = n;
    long long dec_value = 0;
    int base = 1;
    int len = num.length();
    for (int i = len - 1; i >= 0; i--) {
      if (num[i] == '1') dec_value += base;
      base = base * 2;
    }
    return dec_value;
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  bool isPowerOfTwo(int n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  bool isPerfectSquare(long long x) {
    if (x >= 0) {
      long long sr = sqrt(x);
      return (sr * sr == x);
    }
    return false;
  }
  long long moduloMultiplication(long long a, long long b, long long mod) {
    long long res = 0;
    a %= mod;
    while (b) {
      if (b & 1) res = (res + a) % mod;
      b >>= 1;
    }
    return res;
  }
  long long powermod(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  void asquare() {
    long long n, k, x, y, s = 0;
    cin >> n >> k;
    long long arr[n];
    unordered_map<long long, long long> mpp;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      mpp[arr[i]] = 1;
    }
    if (n == 1) {
      cout << k + 1;
    } else if (k < 9) {
      cout << k + 1;
    } else {
      k++;
      for (int i = 0; i < n - 1; i++) {
        if (k > 0) {
          x = (long long)pow(10, arr[i + 1] - arr[i]);
          x--;
          y = min(x, k);
          s += (pow(10, arr[i])) * y;
          k -= y;
        } else
          break;
      }
      if (k > 0) {
        y = k;
        s += (pow(10, arr[n - 1])) * y;
      }
      cout << s;
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      asquare();
    }
    return 0;
  }
  