#include <bits/stdc++.h>
using namespace std;
int main() {
  int test;
  cin >> test;
  while (test--) {
    int bun, beef, chicks, hp, cp, count1 = 0, count2 = 0;
    cin >> bun >> beef >> chicks >> hp >> cp;
    bun /= 2;
    if (hp >= cp) {
      count1 = min(bun, beef);
      bun -= count1;
      if (bun != 0) {
        count2 = min(bun, chicks);
      }
    } else {
      count2 = min(bun, chicks);
      bun -= count2;
      if (bun != 0) {
        count1 = min(bun, beef);
      }
    }
    cout << count1 * hp + count2 * cp << endl;
  }
}

### Prompt

Please formulate 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 test;
  cin >> test;
  while (test--) {
    int bun, beef, chicks, hp, cp, count1 = 0, count2 = 0;
    cin >> bun >> beef >> chicks >> hp >> cp;
    bun /= 2;
    if (hp >= cp) {
      count1 = min(bun, beef);
      bun -= count1;
      if (bun != 0) {
        count2 = min(bun, chicks);
      }
    } else {
      count2 = min(bun, chicks);
      bun -= count2;
      if (bun != 0) {
        count1 = min(bun, beef);
      }
    }
    cout << count1 * hp + count2 * cp << endl;
  }
}

```