#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int b, p, c, h1, h2;
    cin >> b >> p >> c >> h1 >> h2;
    if (b == 0 || (p == 0 && c == 0)) {
      cout << 0 << endl;
    } else if (p == 0) {
      int mx = -1;
      for (int i = 1; i <= c; i++) {
        if ((i * 2) <= b) {
          mx = max(mx, (i * h2));
        } else {
          break;
        }
      }
      cout << mx << endl;
    } else if (c == 0) {
      int mx = -1;
      for (int i = 1; i <= p; i++) {
        if ((i * 2) <= b) {
          mx = max(mx, (i * h1));
        } else {
          break;
        }
      }
      cout << mx << endl;
    } else {
      int mx = -1;
      for (int i = 0; i <= p; i++) {
        for (int j = 0; j <= c; j++) {
          if ((i + j) * 2 <= b) {
            mx = max(mx, (h1 * i) + (h2 * j));
          }
        }
      }
      cout << mx << endl;
    }
  }
  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 t;
  cin >> t;
  while (t--) {
    int b, p, c, h1, h2;
    cin >> b >> p >> c >> h1 >> h2;
    if (b == 0 || (p == 0 && c == 0)) {
      cout << 0 << endl;
    } else if (p == 0) {
      int mx = -1;
      for (int i = 1; i <= c; i++) {
        if ((i * 2) <= b) {
          mx = max(mx, (i * h2));
        } else {
          break;
        }
      }
      cout << mx << endl;
    } else if (c == 0) {
      int mx = -1;
      for (int i = 1; i <= p; i++) {
        if ((i * 2) <= b) {
          mx = max(mx, (i * h1));
        } else {
          break;
        }
      }
      cout << mx << endl;
    } else {
      int mx = -1;
      for (int i = 0; i <= p; i++) {
        for (int j = 0; j <= c; j++) {
          if ((i + j) * 2 <= b) {
            mx = max(mx, (h1 * i) + (h2 * j));
          }
        }
      }
      cout << mx << endl;
    }
  }
  return 0;
}

```