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.text.*;
import java.math.*;
import java.util.regex.*;

public class Main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static PrintWriter pw = new PrintWriter(System.out);
    public static String line;
    public static StringTokenizer st;
    public static StringBuilder out;

    public static void main(String[] args) throws Exception{
        String s = br.readLine();
        TreeMap<Character, Character> d = new TreeMap<Character, Character>();
        d.put(']', '[');
        d.put(')', '(');
        d.put('}', '{');
        d.put('>', '<');
        Stack<Character> S = new Stack<Character>();
        int total = 0;
        boolean f = true;
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(c == '[' || c == '<' || c == '(' || c == '{'){
                S.push(c);
            } else{
                // System.out.println(S.peek());
                if(S.isEmpty()){
                    f = false;
                    break;
                } else if(S.peek() == d.get(c)){
                    S.pop();
                } else{
                    total += 1;
                    S.pop();
                }
            }
        }
        // pw.print(S);
        if(!S.isEmpty()){
            f = false;
        }
        if(f){
            pw.print(total + "\n");
        } else{
            pw.print("Impossible\n");
        }
        pw.close();
        br.close();
    }
}