#include <bits/stdc++.h>
using namespace std;
int64_t fexp(int64_t a, int64_t b) {
  int64_t res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    a[i] = fexp((int64_t)10, (int64_t)a[i]);
  }
  vector<int64_t> b(n, LLONG_MAX);
  for (int i = 1; i < n; i++) {
    b[i - 1] = a[i] / a[i - 1];
    b[i - 1]--;
  }
  int64_t sum = 0;
  int64_t ans = 0;
  for (int i = 0; i < n; i++) {
    if (b[i] == LLONG_MAX) {
      int64_t need = (k + 1) - sum;
      ans += need * a[i];
      break;
    }
    if (sum > k - b[i]) {
      int64_t need = (k + 1) - sum;
      ans += need * a[i];
      break;
    } else {
      sum += b[i];
      ans += b[i] * a[i];
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int64_t fexp(int64_t a, int64_t b) {
  int64_t res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    a[i] = fexp((int64_t)10, (int64_t)a[i]);
  }
  vector<int64_t> b(n, LLONG_MAX);
  for (int i = 1; i < n; i++) {
    b[i - 1] = a[i] / a[i - 1];
    b[i - 1]--;
  }
  int64_t sum = 0;
  int64_t ans = 0;
  for (int i = 0; i < n; i++) {
    if (b[i] == LLONG_MAX) {
      int64_t need = (k + 1) - sum;
      ans += need * a[i];
      break;
    }
    if (sum > k - b[i]) {
      int64_t need = (k + 1) - sum;
      ans += need * a[i];
      break;
    } else {
      sum += b[i];
      ans += b[i] * a[i];
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```