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 Kattist1 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        ArrayList<Integer> pop = new ArrayList<>();
        String input = in.next();
        int p = 0;
        for (int i = 0; i < input.length(); i++){
            if ('<' == input.charAt(i) || '{'== input.charAt(i) || '['== input.charAt(i) || '('== input.charAt(i)){
                p++;
            }
            else
            {
                p--;
            }
            if (p <0){
                break;
            }
        }
        if (p != 0){
            System.out.println("Impossible");
        }
        else{
            for (int i = 0; i < input.length(); i++){
                switch(input.charAt(i)){
                    case ')':{if (pop.get(pop.size()-1) != 1)
                                p++;
                              pop.remove(pop.size()-1);
                              break;}
                    case '}':{if (pop.get(pop.size()-1) != 2)
                                p++;
                              pop.remove(pop.size()-1);
                              break;}
                    case ']':{if (pop.get(pop.size()-1) != 3)
                                p++;
                              pop.remove(pop.size()-1);
                              break;}
                    case '>':{if (pop.get(pop.size()-1) != 4)
                                p++;
                              pop.remove(pop.size()-1);
                              break;}
                    case '(':{pop.add(1);break;}
                    case '{':{pop.add(2);break;}
                    case '[':{pop.add(3);break;}
                    case '<':{pop.add(4);break;}
                }
            }
            System.out.println(p);
        }
    }
}