CodeForces 612C

Replace To Make Regular Bracket Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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 s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2,(s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

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 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

input

[<}){}

output

2

input

{()}[]

output

0

input

]]

output

Impossible

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
char s[1000010];
bool cmp(char a,char b)
{
	 if((a=='['&&b==']')||(a=='('&&b==')')||(a=='{'&&b=='}')||(a=='<'&&b=='>'))
	 return true;
	 else
	 return false;
}
int main()
{
	while(scanf("%s",s)!=EOF)
	{
		stack<char> sta;
		int len=strlen(s);
		int k=0;
		bool flag=true;
		for(int i=0;i<len;i++)
		{
			if(s[i]=='<'||s[i]=='['||s[i]=='{'||s[i]=='(')
			sta.push(s[i]);
			else if(s[i]=='>'||s[i]=='}'||s[i]==']'||s[i]==')')
			{
				if(sta.empty())
				{
					flag=false;
					break;
				}
				else if(!cmp(sta.top(),s[i]))
				{
					k++;	
				}
				sta.pop();
			}
		}
		if(!sta.empty())
		flag=false;
		if(flag)
		printf("%d\n",k);
		else
		printf("Impossible\n");
	}
	return 0;
}