#include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
  int t;
  cin >> t;
  while (t > 0) {
    int b, p, f;
    cin >> b >> p >> f;
    int h, c;
    cin >> h >> c;
    int ans = 0;
    if (h < c) {
      swap(h, c);
      swap(p, f);
    }
    if (p * 2 < b) {
      ans += h * p;
      b -= p * 2;
      if (f * 2 < b) {
        ans += c * f;
      } else
        ans += b / 2 * c;
    } else {
      ans = b / 2 * h;
    }
    cout << ans << endl;
    t--;
  }
  return 0;
}

### 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 argc, const char* argv[]) {
  int t;
  cin >> t;
  while (t > 0) {
    int b, p, f;
    cin >> b >> p >> f;
    int h, c;
    cin >> h >> c;
    int ans = 0;
    if (h < c) {
      swap(h, c);
      swap(p, f);
    }
    if (p * 2 < b) {
      ans += h * p;
      b -= p * 2;
      if (f * 2 < b) {
        ans += c * f;
      } else
        ans += b / 2 * c;
    } else {
      ans = b / 2 * h;
    }
    cout << ans << endl;
    t--;
  }
  return 0;
}

```