#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, b, p, f;
  int h, c;
  scanf("%d", &t);
  while (t--) {
    int cost = 0;
    scanf("%d %d %d %d %d", &b, &p, &f, &h, &c);
    int maxi = max(h, c);
    if (b > 1) {
      if (maxi == c) {
        if (b >= f * 2) {
          b -= f * 2;
          cost += f * c;
        } else if (b < f * 2) {
          cost += b / 2 * c;
          b -= (b / 2) * 2;
        }
        if (b >= p * 2) {
          b -= p * 2;
          cost += p * h;
        } else if (b < p * 2) {
          cost += b / 2 * h;
          b -= (b / 2) * 2;
        }
      } else if (maxi == h) {
        if (b >= p * 2) {
          b -= p * 2;
          cost += p * h;
        } else if (b < p * 2) {
          cost += b / 2 * h;
          b -= (b / 2) * 2;
        }
        if (b >= f * 2) {
          b -= f * 2;
          cost += f * c;
        } else if (b < f * 2) {
          cost += b / 2 * c;
          b -= (b / 2) * 2;
        }
      }
    }
    printf("%d\n", cost);
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 t, b, p, f;
  int h, c;
  scanf("%d", &t);
  while (t--) {
    int cost = 0;
    scanf("%d %d %d %d %d", &b, &p, &f, &h, &c);
    int maxi = max(h, c);
    if (b > 1) {
      if (maxi == c) {
        if (b >= f * 2) {
          b -= f * 2;
          cost += f * c;
        } else if (b < f * 2) {
          cost += b / 2 * c;
          b -= (b / 2) * 2;
        }
        if (b >= p * 2) {
          b -= p * 2;
          cost += p * h;
        } else if (b < p * 2) {
          cost += b / 2 * h;
          b -= (b / 2) * 2;
        }
      } else if (maxi == h) {
        if (b >= p * 2) {
          b -= p * 2;
          cost += p * h;
        } else if (b < p * 2) {
          cost += b / 2 * h;
          b -= (b / 2) * 2;
        }
        if (b >= f * 2) {
          b -= f * 2;
          cost += f * c;
        } else if (b < f * 2) {
          cost += b / 2 * c;
          b -= (b / 2) * 2;
        }
      }
    }
    printf("%d\n", cost);
  }
}

```