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 __future__ import division
from sys import stdin, stdout
# from fractions import gcd
# from math import *
# from operator import mul
# from functools import reduce
# from copy import copy
from collections import deque, defaultdict, Counter

rstr = lambda: stdin.readline().strip()
s, stk, ans = rstr(), [], 0
for i in s:
    if i in ['[', '{', '<', '(']:
        stk.append(i)
    else:
        if not stk:
            print('Impossible')
            exit()
        else:
            lst = stk.pop()
            if lst == '{' and i != '}':
                ans += 1
            elif lst == '[' and i != ']':
                ans += 1
            elif lst == '<' and i != '>':
                ans += 1
            elif lst == '(' and i != ')':
                ans += 1
                
print(ans if not stk else 'Impossible')
