#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
int main() {
  unsigned int t;
  cin >> t;
  while (t--) {
    int b, p, f, h, c;
    scanf("%d", &b);
    scanf("%d", &p);
    scanf("%d", &f);
    scanf("%d", &h);
    scanf("%d", &c);
    int prof = 0;
    if (h >= c) {
      if (b >= 2 * p) {
        prof = prof + p * h;
        b = b - 2 * p;
      } else {
        prof = prof + (b / 2) * h;
        b = b - 2 * (b / 2);
      }
      if (b > 1) {
        if (b >= 2 * f) {
          prof = prof + f * c;
          b = b - 2 * f;
        } else {
          prof = prof + (b / 2) * c;
          b = b - 2 * (b / 2);
        }
      }
    } else {
      if (b >= 2 * f) {
        prof = prof + f * c;
        b = b - 2 * f;
      } else {
        prof = prof + (b / 2) * c;
        b = b - 2 * (b / 2);
      }
      if (b > 1) {
        if (b >= 2 * p) {
          prof = prof + p * h;
          b = b - 2 * f;
        } else {
          prof = prof + (b / 2) * h;
          b = b - 2 * (b / 2);
        }
      }
    }
    cout << prof << endl;
  }
}

### Prompt

Generate 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;
const double pi = acos(-1.0);
int main() {
  unsigned int t;
  cin >> t;
  while (t--) {
    int b, p, f, h, c;
    scanf("%d", &b);
    scanf("%d", &p);
    scanf("%d", &f);
    scanf("%d", &h);
    scanf("%d", &c);
    int prof = 0;
    if (h >= c) {
      if (b >= 2 * p) {
        prof = prof + p * h;
        b = b - 2 * p;
      } else {
        prof = prof + (b / 2) * h;
        b = b - 2 * (b / 2);
      }
      if (b > 1) {
        if (b >= 2 * f) {
          prof = prof + f * c;
          b = b - 2 * f;
        } else {
          prof = prof + (b / 2) * c;
          b = b - 2 * (b / 2);
        }
      }
    } else {
      if (b >= 2 * f) {
        prof = prof + f * c;
        b = b - 2 * f;
      } else {
        prof = prof + (b / 2) * c;
        b = b - 2 * (b / 2);
      }
      if (b > 1) {
        if (b >= 2 * p) {
          prof = prof + p * h;
          b = b - 2 * f;
        } else {
          prof = prof + (b / 2) * h;
          b = b - 2 * (b / 2);
        }
      }
    }
    cout << prof << endl;
  }
}

```