Problem: 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

Solution (in PYTHON3):

Solution:
opening = set(['(','<','[','{'])
closing = set([')','>',']','}'])
 
def im_noob(s):
    o = c = boom = 0
    for i in s:
        if i in opening:
            o+=1
            boom+=1
        else:
            c+=1
            boom-=1
 
        if boom<0:
            return "Impossible"
    
    if o!=c:
        return "Impossible"
    if boom!=0:
        return "Impossible"
 
    ans = 0
    #square = 0
    #angular = 0
    #curly = 0
    #curve = 0
    stack = []
    for i in s:
        if i in opening:
            if i == '[':
                stack.append(']')
 
            elif i == '<':
                stack.append('>')
 
            elif i == '{':
                stack.append('}')
            
            elif i == '(':
                stack.append(')')
        
        else:
            if i!=stack[len(stack)-1]:
                ans+=1
                stack.pop()
            else:
                stack.pop()
            
 
    return ans
 
 
    #l = [square, angular, curly, curve]
    #print(*l)
    #return sum([x for x in l if x>=0])
 
s = input()
print(im_noob(s))