#include <bits/stdc++.h>
using namespace std;
int a[10];
long long v[10];
long long pw[12];
int len(long long x) {
  int ans = 0;
  while (x) {
    ans++;
    x /= 10;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  pw[0] = 1;
  for (int i = 1; i <= 10; i++) pw[i] = pw[i - 1] * 10;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 1; i < n; i++) {
      v[i] = min(k, pw[a[i] - a[i - 1]] - 1);
      k -= v[i];
    }
    bool flag = false;
    if (k != 0) {
      cout << k;
      flag = true;
    }
    for (int i = n - 2; i >= 0; i--) {
      if (flag)
        for (int j = 1; j <= a[i + 1] - a[i] - len(v[i + 1]); j++) cout << "0";
      if (v[i + 1] != 0) {
        cout << v[i + 1];
        flag = true;
      }
    }
    cout << "\n";
  }
}

### Prompt

Create a solution in CPP for 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;
int a[10];
long long v[10];
long long pw[12];
int len(long long x) {
  int ans = 0;
  while (x) {
    ans++;
    x /= 10;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  pw[0] = 1;
  for (int i = 1; i <= 10; i++) pw[i] = pw[i - 1] * 10;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 1; i < n; i++) {
      v[i] = min(k, pw[a[i] - a[i - 1]] - 1);
      k -= v[i];
    }
    bool flag = false;
    if (k != 0) {
      cout << k;
      flag = true;
    }
    for (int i = n - 2; i >= 0; i--) {
      if (flag)
        for (int j = 1; j <= a[i + 1] - a[i] - len(v[i + 1]); j++) cout << "0";
      if (v[i + 1] != 0) {
        cout << v[i + 1];
        flag = true;
      }
    }
    cout << "\n";
  }
}

```