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.io.PrintWriter;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class Main {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        Simba solver = new Simba();
        solver.solve(s, out);

        s.close();
        out.close();
    }

}

class Simba {

    public void solve(Scanner sc, PrintWriter out) {
        /*
         * System.out.println("asdasd"); String s =
         * ")((()))(((())))())()()()()("; String s2 = "((((((()))))))";
         */
        String s2 = "[]{[]({)([";
        findValidParentheses(sc.nextLine());
        // findValidParentheses(s2);

    }

    private void findValidParentheses(String ss) {
        Stack<Character> st = new Stack<>();

        HashMap<Character, Character> hm = new HashMap<>();

        hm.put('<', '>');
        hm.put('(', ')');
        hm.put('{', '}');
        hm.put('[', ']');

        int counter = 0;
        boolean a = true;
        for (int i = 0; i < ss.length(); i++) {

            if (hm.containsKey(ss.charAt(i))) {
                st.push(ss.charAt(i));
            } else {
                if (st.isEmpty()) {
                    System.out.println("Impossible");
                    a = false;
                    break;
                } else if (hm.get(st.pop()) != ss.charAt(i)) {
                    counter++;
                }
            }
        }
        if (!st.isEmpty()) {
            System.out.println("Impossible");
        } else if (a) {

            System.out.println(counter);
        }
    }
}