You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

Input

[&lt;}){}


Output

2

Input

{()}[]


Output

0

Input

]]


Output

Impossible
s = raw_input()


def isClose(c):
    if c == ")":
        return True
    elif c == ">":
        return True
    elif c == "}":
        return True
    elif c == "]":
        return True
    else:
        return False


def sameKind(c1, c2):
    if c1 == "(" and c2 == ")":
        return True
    elif c1 == "<" and c2 == ">":
        return True
    elif c1 == "{" and c2 == "}":
        return True
    elif c1 == "[" and c2 == "]":
        return True
    else:
        return False


def solve():
    stack = []
    ans = 0

    for i in range(len(s)):
        if not isClose(s[i]):
            stack.append(s[i])
        else:
            if len(stack) == 0:
                return "Impossible"
            else:
                c = stack.pop()
                if not sameKind(c, s[i]):
                    ans += 1

    if len(stack) == 0:
        return ans
    else:
        return "Impossible"


print solve()
