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.*;
import java.util.*;
import java.lang.*;


public class Main {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer tokenizer;
    public static String tokenizer() throws Exception {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return (tokenizer.nextToken());
    }
    public static int nextInt() throws Exception {
        return (Integer.parseInt(tokenizer()));
    }
    public static void main(String[] args) throws Exception {
        Stack<Character> set = new Stack<Character>();
        String s;
        Character a, b;
        int n = 0,nn=0;

        s = reader.readLine();
        if (s.length() % 2 == 0) {

            for (int i = 0; i < s.length(); i++) {
                if (set.isEmpty())
                {
                    b = s.charAt(i);
                    if (b == '<' || b == '(' || b == '[' || b == '{')
                        set.push(s.charAt(i));
                    else
                    {
                        System.out.println("Impossible");
                        nn++;
                        break;
                    }
                }
                else {
                    a=set.peek();
                    b = s.charAt(i);
                    if (b == '<' || b == '(' || b == '[' || b == '{')
                        set.push(b);
                    else {
                        if (a=='<'&&b=='>')
                            set.pop();
                        else if (a=='['&&b==']')
                            set.pop();
                        else if (a=='{'&&b=='}')
                            set.pop();
                        else if (a=='('&&b==')')
                            set.pop();
                        else if (b==']'||b=='>'||b==')'||b=='}')
                        {
                            set.pop();
                            n++;
                        }
                        else {
                            System.out.println("Impossible");
                            nn++;
                            break;
                        }
                    }
                }
            }
            if (nn==0&&set.isEmpty())System.out.println(n);
            else if (nn==0)System.out.println("Impossible");

        } else System.out.println("Impossible");
    }
}