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
def search(s):
    i=0
    while(i<len(s)-1):
        if s[i] in o and s[i+1] in c:
            return i
        i+=1
    return len(s)

s=raw_input()
o=['(','[','{','<']
c=[')',']','}','>']
stack=[]

t=0
for i in range(len(s)):
    if s[i] in o: stack.append(s[i])
    if s[i] in c:
        if len(stack)==0:
            print('Impossible')
            quit()
        if o.index(stack[-1])!=c.index(s[i]):
            t+=1
        stack.pop()
#while(len(s)>0):
#    i=search(s)
#    if i==len(s):
#        print 'Impossible'
#        quit()
#    if (o.index(s[i])!=c.index(s[i+1])): t+=1
#    s=s[:i]+s[i+2:]
if len(stack)!=0:
    print 'Impossible'
    quit()

print t
