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
inputCopy
3
15 2 3
5 10
7 5 2
10 12
1 100 100
100 100
outputCopy
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.
首先看价格，先保证高价食物，然后计算剩余食材能卖出多少钱。
代码：

#include<bits/stdc++.h>
int main()
{
	int t, b, p, f, r = 0, c, h;
	scanf("%d", &t);
	while (t--) 
	{
		scanf("%d %d %d", &b, &p, &f);
		scanf("%d %d", &h, &c);
		if (h > c) 
		{
			if ((b / 2) >= p) 
			{
				r = p * h;
				b = (b / 2) - p;
				if (b >= f)
				{
					r = r + f * c;
				}
				else
				{
					r = r + b * c;
				}
			}
			else 
			{
				r = (b / 2) * h;
			}
		}
		else 
		{
			if ((b / 2) >= f) 
			{
				r = f * c;
				b = (b / 2) - f;
				if (b >= p)
				{
					r = r + p * h;
				}
				else
				{
					r = r + b * h;
				}
			}
			else 
			{
				r = (b / 2) * c;
			}
 
		}
		printf("%d\n", r);
	}
	return 0;
}