栈（括号匹配）

C. 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

Copy

[<}){}

output

Copy

2

input

Copy

{()}[]

output

Copy

0

input

Copy

]]

output

Copy

Impossible

#include
#include
#include
using namespace std;
char str[1000005];
int main()
{
	stacks;
	int i,l,flag,ans;
	while(cin>>str)
	{
		l=strlen(str);
		flag=1;ans=0;
		for(i=0;i')
					{
						if(s.top()!='<')
						ans++;
					}
					else if(str[i]=='}')
					{
						if(s.top()!='{')
						ans++;
					}
					else if(str[i]==')')
					{
						if(s.top()!='(')
						ans++;
					}
					else if(str[i]==']')
					{
						if(s.top()!='[')
						ans++;
					}
					s.pop();
				}
			}
		}
		if(!s.empty()||flag==0) cout<<"Impossible"<