#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  vector<vector<int>> v(t, vector<int>(5, 0));
  for (int i = 0; i < t; i++) {
    cin >> v[i][0] >> v[i][1] >> v[i][2];
    cin >> v[i][3] >> v[i][4];
  }
  for (int i = 0; i < t; i++) {
    int b = v[i][0];
    int p = v[i][1];
    int f = v[i][2];
    int h = v[i][3];
    int c = v[i][4];
    int c1 = 0;
    int c2 = 0;
    int t = 0;
    if (h > c) {
      if (b > 2 * p) {
        c1 = p * h;
        if (b - 2 * p > 2 * f)
          c2 = f * c;
        else
          c2 = ((b - 2 * p) / 2) * c;
      } else
        c1 = (b / 2) * h;
      t = c1 + c2;
      cout << t << endl;
    } else {
      if (b > 2 * f) {
        c1 = f * c;
        if (b - 2 * f > 2 * p)
          c2 = p * h;
        else
          c2 = ((b - 2 * f) / 2) * h;
      } else
        c1 = (b / 2) * c;
      t = c1 + c2;
      cout << t << endl;
    }
  }
}

### 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;
  vector<vector<int>> v(t, vector<int>(5, 0));
  for (int i = 0; i < t; i++) {
    cin >> v[i][0] >> v[i][1] >> v[i][2];
    cin >> v[i][3] >> v[i][4];
  }
  for (int i = 0; i < t; i++) {
    int b = v[i][0];
    int p = v[i][1];
    int f = v[i][2];
    int h = v[i][3];
    int c = v[i][4];
    int c1 = 0;
    int c2 = 0;
    int t = 0;
    if (h > c) {
      if (b > 2 * p) {
        c1 = p * h;
        if (b - 2 * p > 2 * f)
          c2 = f * c;
        else
          c2 = ((b - 2 * p) / 2) * c;
      } else
        c1 = (b / 2) * h;
      t = c1 + c2;
      cout << t << endl;
    } else {
      if (b > 2 * f) {
        c1 = f * c;
        if (b - 2 * f > 2 * p)
          c2 = p * h;
        else
          c2 = ((b - 2 * f) / 2) * h;
      } else
        c1 = (b / 2) * c;
      t = c1 + c2;
      cout << t << endl;
    }
  }
}

```