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
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2015 vivek9ru <vivek9ru@Vivek9RU>
#
# Distributed under terms of the MIT license.

from math import acos,asin,atan,sqrt,pi,sin,cos,tan,ceil,floor
import re
from operator import itemgetter as ig

dic={'[':1,'<':2,'{':3,'(':4,']':-1,'>':-2,'}':-3,')':-4}
s=raw_input()
st=[0]*len(s)
le=0
c=0
for i in range(len(s)):
	if le==0:
		st[le]=dic[s[i]]
		le+=1
	else:
		if st[le-1]>0 and dic[s[i]]<0:
			if st[le-1]+dic[s[i]]!=0:
				c+=1
			le-=1
		else:
			st[le]=dic[s[i]]
			le+=1
if le!=0:
	print "Impossible"
else:
	print c
exit(0)
