Yogurt Sale
The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?
In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.
The first line contains a single integer $$$t$$$ ($$$1 \le t \le {10}^{4}$$$) — the number of test cases.The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.
For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".
t = int(input())
output = []
for i in range(t):
    nab = [int(i) for i in input().split()]
    n = nab[0]
    a = nab[1]
    b = nab[2]
    if b < 2 * a:
        price = n // 2 * b + n % 2 * a
    else:
        price = a * n
    output.append(price)
for i in output:
    print(i)
def min_cost(t, cases):
    for case in cases:
        (n, a, b) = case
        if n % 2 == 0:
            cost = min(n * a, n // 2 * b)
        else:
            cost = min(n * a, (n - 1) // 2 * b + a)
        print(cost)
t = int(input())
cases = []
for _ in range(t):
    (n, a, b) = map(int, input().split())
    cases.append((n, a, b))
min_cost(t, cases)
def min_cost(t, cases):
    for case in cases:
        (n, a, b) = case
        if a * 2 <= b:
            print(n * a)
        else:
            pairs = n // 2
            remaining = n % 2
            print(pairs * b + remaining * a)
t = int(input())
cases = []
for _ in range(t):
    (n, a, b) = map(int, input().split())
    cases.append((n, a, b))
min_cost(t, cases)