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
s = input()
cnt = 0
closing = ['>','}',']',')']
opening = ['<','{','[','(']
oc=0
cc=0
for i in s:
    if i in opening:
        oc+=1
    else:
        cc+=1
if s[0] in closing or len(s)&1 or oc!=cc:
    print("Impossible")
else:
    stack = []
    for i in s:
        # print(i,stack)
        if i in closing:
            # print('similar opening',opening[closing.index(i)],stack[-1],opening[closing.index(i)]==s[-1])
            if len(stack)==0:
                print("Impossible")
                exit(0)
            if opening[closing.index(i)]==stack[-1]:
                stack.pop()
            else:
                cnt+=1
                stack.pop()
        elif i in opening:
            stack.append(i)
    #     print(i,stack,cnt)
    # print(stack)
    print(cnt)
 		  	 				 	  			 	  				 				