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;

int main() {
	string s;
	cin>>s;
	int n = s.length(), c=0;
	stack<char> st1;
	char c1, c2;
	int flag = 0;
	for(int i=0; i<n; i++)
	{
		if(s[i] == '(' || s[i] == '[' || s[i] == '{' || s[i] == '<')
		{
			st1.push(s[i]);
		}
		else if(s[i] == ')')
		{
			if(!st1.empty())
			{
				if(st1.top() != '(' && (st1.top() == '{' || st1.top() == '<' || st1.top() == '['))
				{
					c++;
				}
				st1.pop();
			}
			else
			{
				flag = 1;
			}
		}
		else if(s[i] == ']')
		{
			if(!st1.empty())
			{
				if(st1.top() != '[' && (st1.top() == '{' || st1.top() == '<' || st1.top() == '<'))
				{
					c++;
				}
				st1.pop();
			}
			else
			{
				flag = 1;
			}
		}
		else if(s[i] == '}')
		{
			if(!st1.empty())
			{
				if(st1.top() != '{' &&(st1.top() == '>' || st1.top() == '<' || st1.top() == '['))
				{
					c++;
				}
				st1.pop();
			}
			else
			{
				flag = 1;
			}
		}
		else
		{
			if(!st1.empty())
			{
				if(st1.top() != '<' && (st1.top() == '{' || st1.top() == '(' || st1.top() == '['))
				{
					c++;
				}
				st1.pop();
			}
			else
			{
				flag = 1;
			}
		}
	}
	if(flag == 1)
	{
		cout<<"Impossible";
	}
	else
	{
		cout<<c;
	}
	return 0;
}
 		 		 				   	  	      	 	   	