#include <bits/stdc++.h>
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
int main() {
  int buns, p, f, h, c, t;
  scanf("%d", &t);
  int ans[t];
  int count = 0;
  while (t--) {
    int cost = 0;
    scanf("%d%d%d", &buns, &p, &f);
    scanf("%d%d", &h, &c);
    if (buns >= 2) {
      if (c > h) {
        cost = cost + min(buns / 2, f) * c;
        buns = buns - min(buns, f * 2);
        if (buns > 0) {
          cost = cost + min(buns / 2, p) * h;
          buns = buns - min(buns / 2, p);
        }
      } else {
        cost = cost + min(buns / 2, p) * h;
        buns = buns - min(buns, 2 * p);
        if (buns > 0) {
          cost = cost + min(buns / 2, f) * c;
          buns = buns - min(buns / 2, p);
        }
      }
      ans[count++] = cost;
    } else {
      ans[count++] = 0;
    }
  }
  for (int i = 0; i < count; i++) {
    printf("%d\n", ans[i]);
  }
}

### Prompt

Please formulate a CPP solution to 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>
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
int main() {
  int buns, p, f, h, c, t;
  scanf("%d", &t);
  int ans[t];
  int count = 0;
  while (t--) {
    int cost = 0;
    scanf("%d%d%d", &buns, &p, &f);
    scanf("%d%d", &h, &c);
    if (buns >= 2) {
      if (c > h) {
        cost = cost + min(buns / 2, f) * c;
        buns = buns - min(buns, f * 2);
        if (buns > 0) {
          cost = cost + min(buns / 2, p) * h;
          buns = buns - min(buns / 2, p);
        }
      } else {
        cost = cost + min(buns / 2, p) * h;
        buns = buns - min(buns, 2 * p);
        if (buns > 0) {
          cost = cost + min(buns / 2, f) * c;
          buns = buns - min(buns / 2, p);
        }
      }
      ans[count++] = cost;
    } else {
      ans[count++] = 0;
    }
  }
  for (int i = 0; i < count; i++) {
    printf("%d\n", ans[i]);
  }
}

```