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
#!/usr/bin/python

from sys import argv, exit

def rstr():
    return input()

def rint():
    return int(input())

def rints():
    return [int(i) for i in input().split(' ')]

def prnt(*args):
    if '-v' in argv:
        print(*args)

op = ['[', '{', '(', '<']
cl = [']', '}', ')', '>']

oppo = {']': '[', '}': '{', '>': '<', ')': '('}
oppo.update(zip(op, cl))

s = rstr()

opd = {c:0 for c in op}
shtack = []

opi = 0
count = 0
for c in s:
    if c in cl:
        prnt(shtack)
        if not shtack:
            print('Impossible')
            exit(0)
        elif shtack[-1] == oppo[c]:
            shtack.pop()
        else:
            count += 1
            prnt(c, count)
            shtack.pop()
    else:
        shtack.append(c)

if shtack:
    print('Impossible')
    exit(0)

print(count)
