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
n=input()
res=[]
cnt=0
for i in n:
    if i in '(<[{':res.append(i)
    else:
        if not res:print('Impossible');break
        l=res.pop()
        if l+i not in ('[]', '()', '<>', '{}'):cnt+=1
else:
    if res:print('Impossible')
    else:print(cnt)


# a = input()
# l = []
# output = 0
# sl = ('[]', '{}', '<>', '()')
# for i in a:
# 	if i == '[' or i == '{' or i == '<' or i == '(':l.append(i)
# 	else:
# 		if len(l) == 0:
# 			print("Impossible")
# 			exit()
# 		else:
# 			top = l.pop()
# 			if top + i not in sl:
# 				output += 1
# if len(l) == 0:print(output)
# else:print("Impossible")

