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:
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
/**
 *
 * @author HRIDAY
 */
public class Brackets {
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader (System.in));
        String inp = br.readLine();
        Stack bracs = new Stack();
        int count = 0;
        for (int i = 0; i < inp.length(); i++) {
            char brac = inp.charAt(i);
            if(brac == '[' || brac == '{' || brac == '(' || brac == '<') bracs.push(brac);
            else {
                if(bracs.isEmpty()) {
                    System.out.println("Impossible");
                    System.exit(0);
                }
                char p = (char) bracs.pop();
                if( ( p == '[' && brac != ']' ) || ( p == '{' && brac != '}' ) || (p == '(' && brac != ')' ) || (p == '<' && brac != '>') )
                    count++;
                
            }
            
        }
        if(!bracs.isEmpty()) System.out.println("Impossible");
        else System.out.println(count);
    }
}