#include <bits/stdc++.h>
int make(int *, int *);
int sum, num, i, bun, beef, chi, bprice, cprice;
int main(int argc, char *argv[]) {
  int total = 0;
  scanf("%d", &num);
  for (i = 0; i < num; i++) {
    scanf("%d %d %d %d %d", &bun, &beef, &chi, &bprice, &cprice);
    if (bprice > cprice) {
      total += make(&bun, &beef) * bprice;
      total += make(&bun, &chi) * cprice;
    } else {
      total += make(&bun, &chi) * cprice;
      total += make(&bun, &beef) * bprice;
    }
    printf("%d", total);
    total = 0;
    if (i != num - 1) printf("\n");
  }
}
int make(int *bread, int *patty) {
  int tmp;
  if (*bread > *patty * 2) {
    *bread -= *patty * 2;
    tmp = *patty;
    *patty = 0;
    return tmp;
  } else {
    *patty -= *bread / 2;
    tmp = *bread;
    *bread = 0;
    return tmp / 2;
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 make(int *, int *);
int sum, num, i, bun, beef, chi, bprice, cprice;
int main(int argc, char *argv[]) {
  int total = 0;
  scanf("%d", &num);
  for (i = 0; i < num; i++) {
    scanf("%d %d %d %d %d", &bun, &beef, &chi, &bprice, &cprice);
    if (bprice > cprice) {
      total += make(&bun, &beef) * bprice;
      total += make(&bun, &chi) * cprice;
    } else {
      total += make(&bun, &chi) * cprice;
      total += make(&bun, &beef) * bprice;
    }
    printf("%d", total);
    total = 0;
    if (i != num - 1) printf("\n");
  }
}
int make(int *bread, int *patty) {
  int tmp;
  if (*bread > *patty * 2) {
    *bread -= *patty * 2;
    tmp = *patty;
    *patty = 0;
    return tmp;
  } else {
    *patty -= *bread / 2;
    tmp = *bread;
    *bread = 0;
    return tmp / 2;
  }
}

```