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
from collections import defaultdict

def f(n = 20):
    c = 0
    s = []
    for x in xrange(n):
        if c and random.randint(0,1):
            s.append(random.choice(')>]}'))
            c -= 1
        else:
            s.append(random.choice('(<[{'))
            c += 1
    for x in xrange(c):
        s.append(random.choice(')>]}'))
    return ''.join(s)

def pmove(ss):
    ob = '[({<'
    cb = '])}>'
    m = lambda x: {']':'[', ')':'(', '}':'{', '>':'<'}[x]
    cd = defaultdict(int)
    for char in ss:
        if char in ob:
            cd[char] += 1
        else:
            cd[m(char)] -= 1
    return sum((abs(val) for val in cd.values()))/2
            
def solve(bs):
    ob = '[({<'
    cb = '])}>'
    c = 0
    ls = 0
    sbs = []
    for i, char in enumerate(bs):
        if char in ob:
            c+=1
        elif char in cb:
            c-=1
        if c == 0:
            sbs.append(bs[ls:i+1])
            ls = i+1
    if c != 0:
        return 'Impossible'
    else:
        return sum((pmove(ss) for ss in sbs))

def solve2(bs):
    stack = []
    c = 0
    ob = '[({<'
    cb = '])}>'
    m = lambda x: {']':'[', ')':'(', '}':'{', '>':'<'}[x]
    for char in bs:
        if char in ob:
            stack.append(char)
        else:
            if not stack:
                return 'Impossible'
            if stack[-1] != m(char):
                c += 1
            stack.pop()
    if stack:
        return 'Impossible'
    return c

def main():
    lines = stdin.readlines()
    #lines = ['[(){][<]>{}}\n']
    stdout.write(str(solve2(lines[0].strip())))
    
main()
