#include <bits/stdc++.h>
using namespace std;
long long T, n, k;
long long a[15], bb[15], po[15];
long long lp(int t) {
  long long d = 1;
  for (int i = 1; i <= t; ++i) d = d * 10;
  return d;
}
int main() {
  cin >> T;
  for (int ii = 1; ii <= T; ++ii) {
    cin >> n >> k;
    k = k + 1;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      if (i > 1) {
        bb[i - 1] = lp(a[i] - a[i - 1]) - 1;
        po[i - 1] = bb[i - 1] + po[i - 2];
      }
    }
    long long ans = 0;
    for (int d = n - 1; d >= 1; --d)
      if (k > po[d]) {
        ans += lp(a[d + 1]) * (k - po[d]);
        k -= (k - po[d]);
      }
    cout << ans + k << endl;
  }
  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;
long long T, n, k;
long long a[15], bb[15], po[15];
long long lp(int t) {
  long long d = 1;
  for (int i = 1; i <= t; ++i) d = d * 10;
  return d;
}
int main() {
  cin >> T;
  for (int ii = 1; ii <= T; ++ii) {
    cin >> n >> k;
    k = k + 1;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      if (i > 1) {
        bb[i - 1] = lp(a[i] - a[i - 1]) - 1;
        po[i - 1] = bb[i - 1] + po[i - 2];
      }
    }
    long long ans = 0;
    for (int d = n - 1; d >= 1; --d)
      if (k > po[d]) {
        ans += lp(a[d + 1]) * (k - po[d]);
        k -= (k - po[d]);
      }
    cout << ans + k << endl;
  }
  return 0;
}

```