while True:
    try:
        a, b, x = map(int, input().split())
        # print(a, b, x)
        total = []
        # 注意b类竹鼠不存在单个卖的情况
        # 如果a类竹鼠的单价比类b便宜,那就直接全部买a类竹鼠
        if 3 * a <= b or x < 3:
            print(a * x)
        # 如果a类竹鼠的单价比b类贵,那就买b类竹鼠x以内最大3的倍数只,其余用a类竹鼠填充;
        else:
            b_count = x // 3
            a_count = x % 3
            total1 = b * b_count + a * a_count
            # 需要考虑a * a_count的费用比一个b买三只的费用还要贵的情况,这种情况买到的竹鼠数量>x,但是比total1的花费要少
            total2 = b * (b_count + 1)
            print(min(total1, total2))

    except Exception as e:
        break