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 collections import *
from itertools import *
from random import *
from bisect import *
from string import *
from queue import *
from heapq import *
from math import *
from re import *
from sys import *
def fast(): return stdin.readline().strip()
def zzz(): return [int(i) for i in fast().split()]


z, zz = input, lambda: list(map(int, z().split()))
szz, graph, mod, szzz = lambda: sorted(
    zz()), {}, 10**9 + 7, lambda: sorted(zzz())


def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def output(answer): stdout.write(str(answer))


dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]


###########################---Test-Case---#################################
"""

If you Know me , Then you probably don't know me !

"""
###########################---START-CODING---##############################


cnt = 0

arr = fast()
opn = {'(': ')', '<': '>', '[': ']', '{': '}'}
openBract = 0
que = deque()
for i in arr:
    if i in opn:
        openBract += 1
        que.append(opn[i])
    else:
        try:
            x = que.pop()
            if i != x:
                cnt += 1
                que.appendleft(x)
        except:
            print("Impossible")
            exit()


print(cnt if openBract == len(arr) - openBract else "Impossible")
