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
using namespace std;
char a[1000011];
int main()
{
	int l;
	int ans;
	bool dot;
	int left,right;
	while (~scanf ("%s",a))
	{
		l=strlen(a);
		stack<char>s;
		dot=true;
		ans=0;
		int left=0,right=0;
		for (int i=0;i<l;i++)
		{
			if (a[i]=='(' || a[i]=='<' || a[i]=='{' || a[i]=='[')
			{
				s.push(a[i]);
				left++;
			}
			else if (a[i]==')')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='(')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]==']')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='[')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]=='}')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='{')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]=='>')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='<')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
		}
		if (left!=right)
			dot=false;
		if (dot)
			printf ("%d\n",ans);
		else
			printf ("Impossible\n");
	}
	return 0;
}


  			   		 	  	 				   	     	