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

[&lt;}){}


Output

2

Input

{()}[]


Output

0

Input

]]


Output

Impossible
import java.util.*;
public class intro{
	public static void main(String[] args){
		Scanner in=new Scanner(System.in);
		StringBuffer s=new StringBuffer(in.next());
		StringBuffer stek=new StringBuffer();
		char c,d; int ans=0;
		for(int i=0; i<s.length(); i++){
			c=s.charAt(i);
			if (c=='('||c=='['||c=='{'||c=='<'){
				stek.append(c);
			}
			else{
				if (stek.length()==0){
					System.out.println("Impossible");
					return;
				}
				d=stek.charAt(stek.length()-1);
				if (!((d=='('&c==')')|(d=='<'&c=='>')|(d=='['&c==']')|(d=='{'&c=='}')))
					ans++;
				stek.deleteCharAt(stek.length()-1);
			}
		}
		if (stek.length()!=0){
			System.out.println("Impossible");
			return;
		}
		System.out.print(ans);
	}
}