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
word = input()

BRACKETS = [
    ('<', '>'),
    ('{', '}'),
    ('[', ']'),
    ('(', ')'),
]

OPENING = [b[0] for b in BRACKETS]

MATCH = dict(BRACKETS)

IMPOSSIBLE = 'Impossible'

def is_opening(w):
    return w in OPENING

def same_kind(wo, wc):
    return MATCH[wo] == wc

def solve(word):
    temp = []
    result = 0
    for w in word:
        if is_opening(w):
            temp.append(w)
        else:
            if len(temp) == 0:
                return IMPOSSIBLE
            wo = temp.pop()
            if not is_opening(wo):
                return IMPOSSIBLE
            if not same_kind(wo, w):
                result += 1
    if len(temp) != 0:
        return IMPOSSIBLE
    return result
                
print(solve(word))
