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.math.*;

public class Edu4_C {
    public static char getOpposite(char ch) {
        if(ch == ')')return '(';
        if(ch == '}')return '{';
        else if(ch == '>')return '<';
        return '[';
    }
    public static void main(String[] args) throws IOException {
        BufferedInputStream bis = new BufferedInputStream(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(bis));
        String ans = "Impossible";
        int diff = 0, res = 0; 
        String exp = br.readLine().trim();
        Stack<Character> stck = new Stack<Character>();
        for(int i = 0;i <exp.length() && diff >= 0;i++) {
            char ch = exp.charAt(i);
            if(ch == ')' || ch == '}' || ch == ']' || ch == '>') {
                if(stck.isEmpty()) 
                    diff = -1;
                else {
                    char top = stck.pop();
                    if(top != getOpposite(ch))
                        res++;
                    diff--;
                }
            } else {
                stck.push(ch);
                diff++;
            }
                
        }
        System.out.println(((diff != 0))?ans:res);
    }
}