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.*;
public class RBS{
    public static void main(String args[]){
        Stack<Character> pila = new Stack<>();
        Scanner sc = new Scanner(System.in);
        String linea = sc.nextLine();
        boolean imposible = false;
        int count = 0;
        for(int i = 0; i < linea.length(); i++){
            char c = linea.charAt(i);
            switch(c){
                case '}':
                case ')':
                case ']':
                case '>':
                    if(pila.isEmpty()){
                        imposible = true;
                    }else{
                        char cpila = pila.pop();
                        if(!esPar(c, cpila)){
                            count++;
                        }
                    }
                    break;
                default:
                    pila.push(c);
            }
        }
        if(!pila.isEmpty()){
            imposible = true;
        }
        if(imposible){
            System.out.println("Impossible");
        }else{
            System.out.println(count);
        }
    }
    
    private static boolean esPar(char c, char cpila){
        return (cpila == '{' && c == '}') || (cpila == '[' && c == ']')
         || (cpila == '<' && c == '>') || (cpila == '(' && c == ')');
    }
}