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.ArrayDeque;
import java.util.Deque;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String sequence = sc.nextLine();

        Map<Character, Character> inverse = new HashMap<>();
        inverse.put('(', ')');
        inverse.put(')', '(');
        inverse.put('[', ']');
        inverse.put(']', '[');
        inverse.put('<', '>');
        inverse.put('>', '<');
        inverse.put('{', '}');
        inverse.put('}', '{');

        boolean isCorrect = true;
        int moves = 0;
        Deque<Character> deque = new ArrayDeque<>();

        for(int i = 0; i < sequence.length(); i++) {
            Character current = sequence.charAt(i);
            if(current.equals('}') || current.equals('>') || current.equals(')') || current.equals(']')) {
                if(deque.isEmpty()) {
                    isCorrect = false;
                } else {
                    if(deque.pop().equals(inverse.get(current)) == false) {
                        moves = moves + 1;
                    }
                }
            } else {
                deque.push(current);
            }
        }
        if(deque.isEmpty() == false) {
            isCorrect = false;
        }

        if(isCorrect) {
            System.out.println(moves);
        } else {
            System.out.println("Impossible");
        }

    }

}