#include <bits/stdc++.h>
using namespace std;
double PI = 3.141592653589793;
const double EPS = 1e-6;
const int N = 1e6 + 5;
const long long mod = 1e9 + 7;
int oo = 1e9;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int b, p, f, h, c;
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    scanf("%d%d%d%d%d", &b, &p, &f, &h, &c);
    int ans = 0;
    for (int i = 0; i <= 100; i++) {
      int bb = b - 2 * i;
      int pp = p - i;
      if (bb < 0 || pp < 0) break;
      ans = max(ans, i * h + c * min(bb / 2, f));
    }
    printf("%d\n", ans);
  }
}

### Prompt

Your task is to create 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>
using namespace std;
double PI = 3.141592653589793;
const double EPS = 1e-6;
const int N = 1e6 + 5;
const long long mod = 1e9 + 7;
int oo = 1e9;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int b, p, f, h, c;
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    scanf("%d%d%d%d%d", &b, &p, &f, &h, &c);
    int ans = 0;
    for (int i = 0; i <= 100; i++) {
      int bb = b - 2 * i;
      int pp = p - i;
      if (bb < 0 || pp < 0) break;
      ans = max(ans, i * h + c * min(bb / 2, f));
    }
    printf("%d\n", ans);
  }
}

```