#include <bits/stdc++.h>
using namespace std;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    if (n == 1) {
      cout << k + 1 << "\n";
      continue;
    }
    k++;
    bool ok = false;
    long long new_delta = 0, m = 0;
    for (m = 1; m < n && !ok; m++) {
      new_delta = pow(10, a[m] - a[m - 1]) - 1;
      if (new_delta < k)
        k -= new_delta;
      else
        ok = true;
    }
    stringstream ss;
    if (k) ss << k;
    if (m >= 2) {
      if (ok)
        m = a[m - 2];
      else
        m = a[n - 1];
      for (long long i = 0; i < m; i++) ss << "9";
    }
    cout << ss.str() << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    if (n == 1) {
      cout << k + 1 << "\n";
      continue;
    }
    k++;
    bool ok = false;
    long long new_delta = 0, m = 0;
    for (m = 1; m < n && !ok; m++) {
      new_delta = pow(10, a[m] - a[m - 1]) - 1;
      if (new_delta < k)
        k -= new_delta;
      else
        ok = true;
    }
    stringstream ss;
    if (k) ss << k;
    if (m >= 2) {
      if (ok)
        m = a[m - 2];
      else
        m = a[n - 1];
      for (long long i = 0; i < m; i++) ss << "9";
    }
    cout << ss.str() << "\n";
  }
  return 0;
}

```