#include <bits/stdc++.h>
using namespace std;
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b, c;
    cin >> a;
    cin >> b;
    cin >> c;
    int p1, p2;
    cin >> p1;
    cin >> p2;
    int min_no = min(b, c);
    if (a <= 1) {
      cout << 0 << "\n";
      continue;
    }
    if (b * 2 + c * 2 > a) {
      int p = 0;
      if (p2 >= p1) {
        if (c * 2 > a) {
          cout << (a / 2) * p2 << "\n";
          continue;
        } else {
          p += c * p2;
          if ((a - c * 2) <= 1) {
            cout << p << "\n";
            continue;
          }
          p += ((a - c * 2) / 2) * p1;
          cout << p << "\n";
          continue;
        }
      } else if (p1 > p2) {
        if (b * 2 > a) {
          cout << (a / 2) * p1 << "\n";
          continue;
        } else {
          p += b * p1;
          if ((a - b * 2) <= 1) {
            cout << p << "\n";
            continue;
          }
          p += ((a - b * 2) / 2) * p2;
          cout << p << "\n";
          continue;
        }
      }
    } else {
      int p = 0;
      p += (p2 * c) + (p1 * b);
      cout << p << "\n";
      continue;
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

There are two types of burgers in your restaurant — hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet. 

You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve.

You have to answer t independent queries.

Input

The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.

The first line of each query contains three integers b, p and f (1 ≤ b, ~p, ~f ≤ 100) — the number of buns, beef patties and chicken cutlets in your restaurant.

The second line of each query contains two integers h and c (1 ≤ h, ~c ≤ 100) — the hamburger and chicken burger prices in your restaurant.

Output

For each query print one integer — the maximum profit you can achieve.

Example

Input


3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100


Output


40
34
0

Note

In first query you have to sell two hamburgers and three chicken burgers. Your income is 2 ⋅ 5 + 3 ⋅ 10 = 40.

In second query you have to ell one hamburgers and two chicken burgers. Your income is 1 ⋅ 10 + 2 ⋅ 12 = 34.

In third query you can not create any type of burgers because because you have only one bun. So your income is zero.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b, c;
    cin >> a;
    cin >> b;
    cin >> c;
    int p1, p2;
    cin >> p1;
    cin >> p2;
    int min_no = min(b, c);
    if (a <= 1) {
      cout << 0 << "\n";
      continue;
    }
    if (b * 2 + c * 2 > a) {
      int p = 0;
      if (p2 >= p1) {
        if (c * 2 > a) {
          cout << (a / 2) * p2 << "\n";
          continue;
        } else {
          p += c * p2;
          if ((a - c * 2) <= 1) {
            cout << p << "\n";
            continue;
          }
          p += ((a - c * 2) / 2) * p1;
          cout << p << "\n";
          continue;
        }
      } else if (p1 > p2) {
        if (b * 2 > a) {
          cout << (a / 2) * p1 << "\n";
          continue;
        } else {
          p += b * p1;
          if ((a - b * 2) <= 1) {
            cout << p << "\n";
            continue;
          }
          p += ((a - b * 2) / 2) * p2;
          cout << p << "\n";
          continue;
        }
      }
    } else {
      int p = 0;
      p += (p2 * c) + (p1 * b);
      cout << p << "\n";
      continue;
    }
  }
  return 0;
}

```