You are given string *s* consists of opening and closing brackets of four kinds &lt;&gt;, {}, [], (). 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 &lt; by the bracket {, but you can't replace it by ) or &gt;.

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 *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> be a RBS then the strings &lt;*s*<sub class="lower-index">1</sub>&gt;*s*<sub class="lower-index">2</sub>, {*s*<sub class="lower-index">1</sub>}*s*<sub class="lower-index">2</sub>, [*s*<sub class="lower-index">1</sub>]*s*<sub class="lower-index">2</sub>, (*s*<sub class="lower-index">1</sub>)*s*<sub class="lower-index">2</sub> are also RBS.

For example the string "[[(){}]&lt;&gt;]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string *s* RBS.
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 10<sup class="upper-index">6</sup>.
none
[<}){}
2
{()}[]
0
]]
Impossible
>
Impossible
{}
0
{}
0
{]
1
{]
1
{]
1
[]{[]({)([
Impossible
(([{>}{[{[)]]>>]
7
((<>)[]<]><]
3
[[([[(>]>)))[<)>
6
({)[}<)](}
5
(}{)[<][)(]}
6
>}({>]{[}<{<{{)[]]{)]>]]]<(][{)<<<{<<)>)()[>{<]]{}<>}}}}(>}<})(][>{((<{<)]}>)))][>[}[])<]){]]][<[)([
Impossible
<<[<{{<([({<<[)<>(]]){})>[](])[)))[[}>]<)>[[>{>>>[<]}<>>)[>]<{)<[><(<]][>(>]>][(<][{]}(()<[()[>><<])<]})]<]}{)
45
[<<{{((}[}<<)<)>})(][{>}})((>)<[)[>}[})[)>()[()[((}<<(>)<>](<>(}[>})[[[{)<}<<(}{>>}[<([[])<><)]<{>}[>>>{({>)}]})>)
42
(<[([(<({>(}{]>[(})])}])()<<}{]{[>]>(>>[(>>}[){(}<[{(()]{{<(<{][[{<><{<{)<>>]}}}{)(}{})}[<))>>}((({>){({}{{]}]>>}})>))
43
{(]}<([]<]{>]<{<({{{<>))}[({(}{)[}({>]}}<<)}<]))(<>(>{>{{{)<}({<]<>{)(>[)>{({}<([<[[)]><>{]}}(>]{}<)[})]}]]}]}>}
40
<(([)]<[}>{)>][[(<()({{{>>((]<}<{{<>}><<[)}[[)([[)[)}<]<{(<>){([)><{[([[][({])})<][(}]}>>[){)[({>){(}[}})>}>]>)]}))})>}]
45
)(
Impossible
[[[[[[[[
Impossible
# a=1
# N=int(input())
# def dochisla(a):
#     if a==N:
#         return a
#     else:
#         print(a)
#         return dochisla(a+1)
# print(dochisla(a))
#
# a=int(input())
# b=int(input())
# c=[]
# def dochisla(a):
#     if a==b:
#         return a
#     else:
#         print(a)
#         return dochisla(a+1)
# def otchisla(b):
#    if a>=b:
#        return b
#    else:
#        print(b)
#        return otchisla(b-1)
# print(otchisla(b))
# print(dochisla(a))
#
# def sum(a):
#     if len(a) == 0:
#         return 0
#     else:
#         return a[0] + sum(a[1:])
# print(sum([1, 2, 3, 4, 5]))
#
# def f(n):
#     if n <= 1:
#         return n
#     else:
#         return f(n-1) + f(n-2)
# for i in range(10):
#     print(f(i))
#
# a=int(input("введите число"))
# n=int(input("введите степень"))
# def power(a, n):
#     if n == 0:
#         return a-(a-1)
#     else:
#         r = power(a, n//2)
#         if n % 2 == 0:
#             return r*r
#         else:
#             return r*r*a
# print(power(a,n))
#
# def cifri(n):
#     if n < 10:
#         return n
#     else:
#         return n % 10 + cifri(n // 10)
# n = int(input("Введите число: "))
# r = cifri(n)
# print(f"Сумма цифр равна {r}")
#

# def sum_nested_lists(lst):
#     summ = 0
#     for elem in lst:
#         if elem in list:
#             summ += sum_nested_lists(elem)
#         else:
#             summ += elem
#     return summ
# print(sum_nested_lists([1,2,3,4,5]))
s = input()
stack = []
replacements = 0

for c in s:
    if c in "({[<":
        stack.append(c)
else:
    if not stack:
        replacements += 1
    elif c == ")" and stack[-1] == "(":
         stack.pop()
    elif c == "}" and stack[-1] == "{":
         stack.pop()
    elif c == "]" and stack[-1] == "[":
         stack.pop()
    elif c == ">" and stack[-1] == "<":
         stack.pop()
    else:
         replacements += 1
if stack==0:
      print("Impossible")
else:
      print(replacements)