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.util.HashMap;
import java.util.Scanner;
import java.util.Stack;

public class CReplaceToMakeRegularBracketSequence {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        final String str = sc.nextLine();
        final Stack<Character> stack = new Stack<>();
        HashMap<Character, Character> brackets = new HashMap<>(4);
        brackets.put('<', '>');
        brackets.put('{', '}');
        brackets.put('[', ']');
        brackets.put('(', ')');

        int changes = 0;
        for (int i = 0; i < str.length(); i++) {
            final char c = str.charAt(i);

            if (brackets.keySet().contains(c)) { // If it is open bracket - just add to stack
                stack.push(c);
            } else { // Otherwise it must be closing bracket - need to check stack
                if(stack.isEmpty() || !brackets.keySet().contains(stack.peek())) {
                    System.out.println("Impossible");
                    return;
                }
                final char cPair = stack.pop();
                if (brackets.get(cPair) != c) {
                    changes++;
                }
            }
        }
        if(!stack.isEmpty()) {
            System.out.println("Impossible");
            return;
        }
        System.out.println(changes);
    }
}