#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, a, profit, d;
  cin >> t;
  int b[t], p[t], f[t], h[t], c[t];
  for (a = 0; a < t; a++) {
    cin >> b[a] >> p[a] >> f[a] >> h[a] >> c[a];
    if (b[a] >= 2 * (p[a] + f[a])) {
      profit = p[a] * h[a] + f[a] * c[a];
    }
    if (b[a] < 2 * (p[a] + f[a])) {
      if (h[a] >= c[a]) {
        if (b[a] % 2 == 1) {
          b[a] = b[a] - 1;
        }
        if (p[a] >= b[a] / 2) {
          profit = h[a] * b[a] / 2;
        }
        if (p[a] < b[a] / 2) {
          d = b[a] / 2 - p[a];
          if (f[a] >= d) {
            profit = h[a] * p[a] + d * c[a];
          }
          if (f[a] < d) {
            profit = h[a] * p[a] + f[a] * c[a];
          }
        }
      }
      if (h[a] < c[a]) {
        if (b[a] % 2 == 1) {
          b[a] = b[a] - 1;
        }
        if (f[a] >= b[a] / 2) {
          profit = c[a] * b[a] / 2;
        }
        if (f[a] < b[a] / 2) {
          d = b[a] / 2 - f[a];
          if (p[a] >= d) {
            profit = c[a] * f[a] + d * h[a];
          }
          if (p[a] < d) {
            profit = c[a] * f[a] + p[a] * h[a];
          }
        }
      }
    }
    cout << profit << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 t, a, profit, d;
  cin >> t;
  int b[t], p[t], f[t], h[t], c[t];
  for (a = 0; a < t; a++) {
    cin >> b[a] >> p[a] >> f[a] >> h[a] >> c[a];
    if (b[a] >= 2 * (p[a] + f[a])) {
      profit = p[a] * h[a] + f[a] * c[a];
    }
    if (b[a] < 2 * (p[a] + f[a])) {
      if (h[a] >= c[a]) {
        if (b[a] % 2 == 1) {
          b[a] = b[a] - 1;
        }
        if (p[a] >= b[a] / 2) {
          profit = h[a] * b[a] / 2;
        }
        if (p[a] < b[a] / 2) {
          d = b[a] / 2 - p[a];
          if (f[a] >= d) {
            profit = h[a] * p[a] + d * c[a];
          }
          if (f[a] < d) {
            profit = h[a] * p[a] + f[a] * c[a];
          }
        }
      }
      if (h[a] < c[a]) {
        if (b[a] % 2 == 1) {
          b[a] = b[a] - 1;
        }
        if (f[a] >= b[a] / 2) {
          profit = c[a] * b[a] / 2;
        }
        if (f[a] < b[a] / 2) {
          d = b[a] / 2 - f[a];
          if (p[a] >= d) {
            profit = c[a] * f[a] + d * h[a];
          }
          if (p[a] < d) {
            profit = c[a] * f[a] + p[a] * h[a];
          }
        }
      }
    }
    cout << profit << endl;
  }
  return 0;
}

```