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
from sys import stdin, stdout

        
class SOLVE:
    def solve(self):
        R = stdin.readline
        #f = open('input.txt');R = f.readline
        W = stdout.write
        
        s = R()[:-1]
        
        brackets, cnt, check = [], 0, 0
        for i in range(len(s)):
            if s[i] in ['(', '<', '{', '[']:
                brackets.append(s[i])
                check +=1
            else:
                if len(brackets) == 0:
                    W("Impossible\n")
                    return 0
                
                if s[i] == ')':
                    if brackets[-1] != '(':
                        cnt += 1
                
                elif s[i] == '>':
                    if brackets[-1] != '<':
                        cnt += 1
                
                elif s[i] == '}':
                    if brackets[-1] != '{':
                        cnt += 1
                
                elif s[i] == ']':
                    if brackets[-1] != '[':
                        cnt += 1
                
                check -= 1
                brackets.pop()
                
            if check < 0:
                W("Impossible\n")
                return 0
            
        if check:
            W("Impossible\n")
            return 0
        
        W('%d\n' % cnt)
        return 0
        
def main():
    s = SOLVE()
    s.solve()
main()