#include <bits/stdc++.h>
int main() {
  int t, b, p, c, x, y, i, k, r, q;
  scanf("%d", &t);
  for (i = 1; i <= t; i++) {
    scanf("%d %d %d %d %d", &b, &p, &c, &x, &y);
    b = b / 2;
    if (x >= y) {
      if (b >= p) {
        q = p * x;
        b = b - p;
      } else {
        q = b * x;
        b = 0;
      }
      if (b >= c)
        r = c * y;
      else
        r = b * y;
    }
    if (y > x) {
      if (b >= c) {
        q = c * y;
        b = b - c;
      } else {
        q = b * y;
        b = 0;
      }
      if (b >= p)
        r = p * x;
      else
        r = b * x;
    }
    k = q + r;
    printf("%d\n", k);
  }
}

### Prompt

Please create a solution in cpp 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>
int main() {
  int t, b, p, c, x, y, i, k, r, q;
  scanf("%d", &t);
  for (i = 1; i <= t; i++) {
    scanf("%d %d %d %d %d", &b, &p, &c, &x, &y);
    b = b / 2;
    if (x >= y) {
      if (b >= p) {
        q = p * x;
        b = b - p;
      } else {
        q = b * x;
        b = 0;
      }
      if (b >= c)
        r = c * y;
      else
        r = b * y;
    }
    if (y > x) {
      if (b >= c) {
        q = c * y;
        b = b - c;
      } else {
        q = b * y;
        b = 0;
      }
      if (b >= p)
        r = p * x;
      else
        r = b * x;
    }
    k = q + r;
    printf("%d\n", k);
  }
}

```