def find_last(char, s):
    last = -1
    for i in range(len(s)):
        if char == s[i]:
            last = i
    return last


def find_first_from(i, char, s):
    # naðe prvi kraj zagrade of pocetka prosle koja je zadnja
    for i in range(i, len(s)):
        if char == s[i]:
            return i
    return -1


def is_float(n):
    try:
        float(n)
        return True
    except ValueError:
        return False


def solve(s):
    s = s.replace("*--", "*").replace("/--", "/").replace("+--", "+").replace("---", "-")
    print(s)

    # jednostavni
    if is_float(s):
        return float(s)
    count, op = 0, None
    for operator in operators:
        count += s.count(operator)
        if s.find(operator) != -1 and op == None:
            op = operator
    if count == 1:
        # 1 operator, a + b oblik
        a = float(s[:s.find(op)])
        b = float(s[s.find(op) + 1:])
        if op == "*":
            return a * b
        elif op == "/":
            return a / b
        elif op == "+":
            return a + b
        elif op == "-":
            return a - b
        else:
            print("error??")
    elif count == 2 and (s.count("+-") + s.count("--") + s.count("/-") + s.count("*-")):
        # ERROR: nekad se moze nagomilati jos vise minusa i onda ne radi, ne da mi se ispravljat
        # ako imamo minus, npr 6*-0.4 kod (20+4)*(20+4)+(((16-3)*5)*(3/5-1))
        # onda obrni predznak
        a = float(s[:s.find(op)])
        b = float(s[s.find(op) + 1:])
        op = op[0]  # uzmi prvi dio operatora
        if op == "*":
            return a * b
        elif op == "/":
            return a / b
        elif op == "+":
            return a + b
        elif op == "-":
            return a - b
        else:
            print("error??")

    # kompleksni
    if s.find("(") != -1:
        # ima zagrade
        start = find_last("(", s)
        kraj = find_first_from(start, ")", s)
        s = s[:start] + str(solve(s[start + 1: kraj])) + s[kraj + 1:]
        return solve(s)

    if s.find("+") != -1:
        a = s[:s.find("+")]
        b = s[s.find("+") + 1:]
        return solve(a) + solve(b)

    # za minus treba gledat sve jer u situaciji tipa: "-5-5" nam ce gledat samo prvi i nema rjesenja
    for i in range(len(s)):
        if s[i] == "-":
            # treba razlikovat je li negativan broj ili minus kao operator
            if i == 0 or s[i - 1].isnumeric() == False:
                pass
            else:
                a = s[:i]
                b = s[i + 1:]
                return solve(a) - solve(b)

    if s.find("*") != -1:
        # ima puta
        # s = a * b
        # a i mogu biti slozeni
        a = s[:s.find("*")]
        b = s[s.find("*") + 1:]
        return solve(a) * solve(b)

    if s.find("/") != -1:
        a = s[:s.find("/")]
        b = s[s.find("/") + 1:]
        return solve(a) / solve(b)

    return s


operators = ["(", ")", "*", "/", "+", "-"]
s = input()
print(solve(s))