#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main(int argc, char **argv) {
  long long t;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> banknotes;
    long long temp;
    long long howMany[10] = {0,     9,      99,      999,      9999,
                             99999, 999999, 9999999, 99999999, 999999999};
    long long values[10] = {1,      10,      100,      1000,      10000,
                            100000, 1000000, 10000000, 100000000, 1000000000};
    vector<long long> notesLeft;
    for (long long i = 0; i < n; i++) {
      cin >> temp;
      banknotes.push_back(temp);
      if (i != 0) {
        notesLeft.push_back(howMany[banknotes[i] - banknotes[i - 1]]);
      }
    }
    long long sol = 0;
    for (long long i = 0; i < notesLeft.size(); i++) {
      if (notesLeft[i] < k) {
        sol += values[banknotes[i]] * notesLeft[i];
        k -= notesLeft[i];
      } else {
        sol += values[banknotes[i]] * k;
        k = 0;
      }
    }
    if (k > 0) {
      sol += values[banknotes[banknotes.size() - 1]] * k;
    }
    cout << sol << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main(int argc, char **argv) {
  long long t;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> banknotes;
    long long temp;
    long long howMany[10] = {0,     9,      99,      999,      9999,
                             99999, 999999, 9999999, 99999999, 999999999};
    long long values[10] = {1,      10,      100,      1000,      10000,
                            100000, 1000000, 10000000, 100000000, 1000000000};
    vector<long long> notesLeft;
    for (long long i = 0; i < n; i++) {
      cin >> temp;
      banknotes.push_back(temp);
      if (i != 0) {
        notesLeft.push_back(howMany[banknotes[i] - banknotes[i - 1]]);
      }
    }
    long long sol = 0;
    for (long long i = 0; i < notesLeft.size(); i++) {
      if (notesLeft[i] < k) {
        sol += values[banknotes[i]] * notesLeft[i];
        k -= notesLeft[i];
      } else {
        sol += values[banknotes[i]] * k;
        k = 0;
      }
    }
    if (k > 0) {
      sol += values[banknotes[banknotes.size() - 1]] * k;
    }
    cout << sol << endl;
  }
  return 0;
}

```