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
import sys
import heapq
import bisect
import math
import random

INF = 10**9+7
OFFLINE = 0
N = 101010
sys.setrecursionlimit(INF)

def fi():
	return int(sys.stdin.readline())

def fi2():
	return map(int, sys.stdin.readline().split())

def fi3():
	return sys.stdin.readline().rstrip()

def fo(*args):
	for s in args:
		sys.stdout.write(str(s)+" ")
	sys.stdout.write("\n")

##
if OFFLINE:
	sys.stdin = open("fin.txt", "r")
	sys.stdout = open("fout.txt", "w")
##



##main

stack = []
s = fi3().rstrip()

res = 0

closing = '}])>'
opening = '<({['

revert = {'}':'{', ')':'(', ']':'[', '>':'<'}

for char in s:

	if char in opening:
		stack.append(char)

	if char in closing:
		if len(stack) == 0:
			fo('Impossible')
			exit(0)

		if stack[-1] != revert[char]:
			res += 1

		k = stack.pop(-1)

if len(stack) != 0:
	fo('Impossible')
	exit(0)

fo(res)