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
'''
Replace To Make Regular Bracket Sequence
'''

S = input()
stack = []
count = 0
length = len(S)
flag = 0

if(length % 2):
    flag = 1
else:
    for i in range(length):
        if S[i] =='<' or S[i] =='(' or S[i] =='{' or S[i] =='[':
            stack.append(S[i])
        elif stack != []:
            #print(S[i])
            if S[i] == '>' and stack.pop() != '<':
                count += 1
            elif S[i] == ')' and stack.pop() != '(':
                count += 1
            elif S[i] == '}' and stack.pop() != '{':
                count += 1
            elif S[i] == ']' and stack.pop() != '[':
                count += 1
            if(flag):
                break
        else:
            flag = 1
            break
if flag != 0 or len(stack) != 0:
    print("Impossible")
else:
    print(count)
				  	 	      	    		 			 	