#include <bits/stdc++.h>
using namespace std;
int main() {
  int b, p, f, i, t, profit = 0, minn = 0, h, c, maxx = 0;
  cin >> t;
  for (i = 1; i <= t; i++) {
    profit = 0;
    cin >> b >> p >> f;
    cin >> h >> c;
    if (b < 2) {
      cout << 0 << endl;
    } else if (h != c) {
      if (h > c) {
        if (b >= (2 * p)) {
          profit = p * h;
          b = (b - 2 * p);
          if (b >= 2) {
            if (b >= (2 * f)) {
              profit += f * c;
              cout << profit << endl;
            } else {
              profit += (b / 2) * c;
              cout << profit << endl;
            }
          } else {
            cout << profit << endl;
          }
        } else {
          profit += (b / 2) * h;
          cout << profit << endl;
        }
      } else if (h < c) {
        if (b >= (2 * f)) {
          profit = f * c;
          b = b - (2 * f);
          if (b >= 2) {
            if (b >= (2 * p)) {
              profit += p * h;
              cout << profit << endl;
            } else {
              profit += (b / 2) * h;
              cout << profit << endl;
            }
          } else {
            cout << profit << endl;
          }
        } else {
          profit += (b / 2) * c;
          cout << profit << endl;
        }
      }
    } else if (h == c) {
      if (b <= 2 * (p + f)) {
        profit = (b / 2) * c;
        cout << profit << endl;
      } else {
        profit = (p + f) * c;
        cout << profit << endl;
      }
    }
  }
}

### Prompt

Create a solution in Cpp for 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 b, p, f, i, t, profit = 0, minn = 0, h, c, maxx = 0;
  cin >> t;
  for (i = 1; i <= t; i++) {
    profit = 0;
    cin >> b >> p >> f;
    cin >> h >> c;
    if (b < 2) {
      cout << 0 << endl;
    } else if (h != c) {
      if (h > c) {
        if (b >= (2 * p)) {
          profit = p * h;
          b = (b - 2 * p);
          if (b >= 2) {
            if (b >= (2 * f)) {
              profit += f * c;
              cout << profit << endl;
            } else {
              profit += (b / 2) * c;
              cout << profit << endl;
            }
          } else {
            cout << profit << endl;
          }
        } else {
          profit += (b / 2) * h;
          cout << profit << endl;
        }
      } else if (h < c) {
        if (b >= (2 * f)) {
          profit = f * c;
          b = b - (2 * f);
          if (b >= 2) {
            if (b >= (2 * p)) {
              profit += p * h;
              cout << profit << endl;
            } else {
              profit += (b / 2) * h;
              cout << profit << endl;
            }
          } else {
            cout << profit << endl;
          }
        } else {
          profit += (b / 2) * c;
          cout << profit << endl;
        }
      }
    } else if (h == c) {
      if (b <= 2 * (p + f)) {
        profit = (b / 2) * c;
        cout << profit << endl;
      } else {
        profit = (p + f) * c;
        cout << profit << endl;
      }
    }
  }
}

```