#include <bits/stdc++.h>
using namespace std;
void smeni(int &a, int &b) {
  int tmp;
  tmp = a;
  a = b;
  b = tmp;
}
int funk(int b, int p, int f, int h, int c) {
  if (b <= 1 || p == 0 || f == 0) return 0;
  int suma = 0;
  if (c > h) {
    smeni(c, h);
    smeni(p, f);
  }
  if (c == h) {
    while (b > 1 && p > 0) {
      suma = suma + h;
      b = b - 2;
      p--;
    }
  } else {
    while (b > 1 && p > 0) {
      suma = suma + h;
      p--;
      b = b - 2;
    }
  }
  if (b - 2 * f > 1) {
    suma = suma + f * c;
  } else {
    while (b > 1 && f > 0) {
      suma = suma + c;
      b = b - 2;
      f--;
    }
  }
  return suma;
}
int main() {
  int t;
  cin >> t;
  int b, p, f;
  int h, c;
  queue<int> red;
  for (int i = 0; i < t; i++) {
    cin >> b >> p >> f;
    cin >> h >> c;
    red.push(funk(b, p, f, h, c));
  }
  while (!red.empty()) {
    cout << red.front() << endl;
    red.pop();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void smeni(int &a, int &b) {
  int tmp;
  tmp = a;
  a = b;
  b = tmp;
}
int funk(int b, int p, int f, int h, int c) {
  if (b <= 1 || p == 0 || f == 0) return 0;
  int suma = 0;
  if (c > h) {
    smeni(c, h);
    smeni(p, f);
  }
  if (c == h) {
    while (b > 1 && p > 0) {
      suma = suma + h;
      b = b - 2;
      p--;
    }
  } else {
    while (b > 1 && p > 0) {
      suma = suma + h;
      p--;
      b = b - 2;
    }
  }
  if (b - 2 * f > 1) {
    suma = suma + f * c;
  } else {
    while (b > 1 && f > 0) {
      suma = suma + c;
      b = b - 2;
      f--;
    }
  }
  return suma;
}
int main() {
  int t;
  cin >> t;
  int b, p, f;
  int h, c;
  queue<int> red;
  for (int i = 0; i < t; i++) {
    cin >> b >> p >> f;
    cin >> h >> c;
    red.push(funk(b, p, f, h, c));
  }
  while (!red.empty()) {
    cout << red.front() << endl;
    red.pop();
  }
  return 0;
}

```