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 a{
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        char[] ar=in.next().toCharArray();

        Stack<Character> stack=new Stack<Character>();
        int count=0;

        for(int i=0;i<ar.length;i++){

            if(ar[i]=='<'||ar[i]=='{'||ar[i]=='['||ar[i]=='(')
                stack.push(ar[i]);

            else{

                if(!stack.isEmpty()){
                    int ascii1=(int)stack.pop();
                    int ascii2=(int)ar[i];
                    if(ascii1+1!=ascii2&&ascii1+2!=ascii2) count++;

                }
                else{
                    System.out.println("Impossible");
                    System.exit(0);
                }
            }
        }

        if(stack.isEmpty())
            System.out.println(count);
        else
            System.out.println("Impossible");

    }
}
