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
'''input
][()()
]]
{()}[]
[<}){}
'''
s = raw_input()
last_bracket = []
balance = 0
answer = 0

def open_bracket(c):
	closed = '])>}'
	opens = '[(<{'
	return opens[closed.find(c)]

for i in s:
	if i == '[' or i == '(' or i == '<' or i == '{':
		last_bracket.append(i)
		balance += 1
	else:
		pop = last_bracket.pop() if len(last_bracket) > 0 else False
		# print pop, i, pop != i
		if pop != False and pop != open_bracket(i):
			answer += 1
		balance -= 1

	if balance < 0:
		break;

print answer if balance == 0 else 'Impossible'
# print balance, answer
# print answer
