Problem: 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

Solution (in JAVA):

Solution:
import java.util.Scanner;
import java.util.Stack;

public class Match612C {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc =new Scanner(System.in);
        String s=sc.nextLine();
        Stack<Character> s1=new Stack();
        Stack<Character> s2=new Stack();
        int replace=0;
        int i=0;
        for( i=0;i<s.length();i++)
        {
            if(s.charAt(i)=='{'||s.charAt(i)=='['||s.charAt(i)=='<'||s.charAt(i)=='(')
            {
                s1.push(s.charAt(i));
                //System.out.println(s1.peek());

            }
            
            else
                {
                if((!s1.isEmpty())&&((s1.peek().equals('{')&&s.charAt(i)=='}')||(s1.peek().equals('(')&&s.charAt(i)==')')||(s1.peek().equals('[')&&s.charAt(i)==']')||(s1.peek().equals('<')&&s.charAt(i)=='>'))){
                
                s1.pop();
                //System.out.println(s1.peek());
            }
            else if(s1.isEmpty()){
                System.out.println("Impossible");
                break;
            }
            else{
                replace++;
                s1.pop();
                //System.out.println(s2.peek());
            }
        }
        }
    
if(i==s.length()){
    if(s1.isEmpty())
    System.out.println(replace);
    else
        System.out.println("Impossible");
        
}
    
        

}

}