#include <bits/stdc++.h>
using namespace std;
long long fastpow(long long base, long long exp) {
  long long ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (ans * base);
    base = (base * base);
    exp /= 2;
  }
  return ans;
}
int main() {
  int t, i;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    for (i = 0; i < n; i++) cin >> v[i];
    long long j;
    vector<long long> a(n);
    for (i = 0; i < n; i++) {
      a[i] = fastpow(10, v[i]);
    }
    k++;
    long long z, ans = 0;
    for (i = 0; i < n - 1; i++) {
      z = min(k, a[i + 1] / a[i] - 1);
      ans += z * a[i];
      k -= z;
      if (k < 0) {
        break;
      }
    }
    if (k > 0) {
      ans += a[n - 1] * k;
    }
    cout << ans << endl;
  }
}

### Prompt

Please create a solution in Cpp to 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 fastpow(long long base, long long exp) {
  long long ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (ans * base);
    base = (base * base);
    exp /= 2;
  }
  return ans;
}
int main() {
  int t, i;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    for (i = 0; i < n; i++) cin >> v[i];
    long long j;
    vector<long long> a(n);
    for (i = 0; i < n; i++) {
      a[i] = fastpow(10, v[i]);
    }
    k++;
    long long z, ans = 0;
    for (i = 0; i < n - 1; i++) {
      z = min(k, a[i + 1] / a[i] - 1);
      ans += z * a[i];
      k -= z;
      if (k < 0) {
        break;
      }
    }
    if (k > 0) {
      ans += a[n - 1] * k;
    }
    cout << ans << endl;
  }
}

```