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/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Jun 22 13:09:30 2020

@author: shailesh
"""
def check_open(char):
    if char in '<{[(':
        return 1
    return 0

def find_type(char):
    if char in '()':
        return 0
    elif char in '[]':
        return 1
    elif char in '{}':
        return 2
    else:
        return 3

s = input()

stack = []
impossible = 0
count = 0

#if not check_open(s[0]):
#    impossible = 1
#else:
#    stack.append(s[0])

for char in s:
    if check_open(char):
        stack.append(char)
    else:
        if len(stack) == 0:
            impossible = True
            break
        if check_open(stack[-1]):
            if find_type(char) != find_type(stack[-1]):
                count +=1
            stack.pop()
        else:
            impossible = True
            break
if len(stack) != 0:
    impossible = True


if impossible:
    print('Impossible')
else:
    print(count)