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
import sys
s=input()
s1=s
rc="{<["
for c in rc:
    s1=s1.replace(c,'(')
rc="}>]"
for c in rc:
    s1=s1.replace(c,')')
ts=''
while ts != s1:
    ts=s1
    s1=s1.replace('()','')
if len(s1) !=0:
    print('Impossible')
    sys.exit()

ts=''
while ts != s:
    ts=s
    s=s.replace('()','')
    s = s.replace('[]', '')
    s = s.replace('<>', '')
    s = s.replace('{}', '')
c=0
stack = []
brackets = {')':'(',']':'[','}':'{','>':'<'}
for char in s:
    if char in brackets.values():  # 如果是正括号,入栈
        stack.append(char)
    elif char in brackets.keys():  # 如果是反括号
        if brackets[char] != stack.pop():  # 如果不匹配弹出的栈顶元素
            c=c+1
print(c)
	  	  		  				 	 	 		  	 					