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
#include<bits/stdc++.h>
using namespace std;
#define LL long long 
const int MAXN =1e6+10;
char s[MAXN];
int main(){
    gets(s);
    stack<char>S; while(!S.empty())  S.pop();
    int cnt=0;int flag=1;
    for(int i=0;s[i];i++){
        if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(') S.push(s[i]);
        else if(s[i]=='>'){
            if(S.empty()) { flag=0;break;}
            if(S.top()=='<') {S.pop();}
            else { cnt++; S.pop();} 
        }else if(s[i]=='}'){
            if(S.empty()) { flag=0;break;}
            if(S.top()=='{') {S.pop();}
            else { cnt++; S.pop();}
        }else if(s[i]==']') {
            if(S.empty()) { flag=0;break;}
            if(S.top()=='[') {S.pop();}
            else { cnt++; S.pop();}
        }else if(s[i]==')'){
            if(S.empty()) { flag=0;break;}
            if(S.top()=='(') {S.pop();}
            else { cnt++; S.pop();}
        }
    }
    if(!flag||!S.empty()) puts("Impossible");
    else printf("%d\n",cnt);
    return 0;
}
 		  	 	  	 			 		  	 		 		  	