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.BufferedInputStream;
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;

/**
 * Created by Morton on 3/26/16.
 */
public class RegularBracketSequence {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        String s = in.next();
        Stack<Character> brackets = new Stack<>();
        int r = 0;  //replacements
        for (char c:s.toCharArray()) {
            if (isClosing(c))
                if (brackets.empty()) {
                    System.out.println("Impossible");
                    System.exit(0);
                } else {
                    if (brackets.pop() != reverse(c)) r++;
                }
            else brackets.push(c);
        }
        if (brackets.empty()) {
            System.out.println(r);
        } else {
            System.out.println("Impossible");
        }
    }

    private static char reverse(char c) {
        switch (c) {
            case '[': return ']';
            case '(': return ')';
            case '{': return '}';
            case '<': return '>';
            case ']': return '[';
            case '}': return '{';
            case ')': return '(';
            case '>': return '<';
            default: throw new IllegalArgumentException("Argument can only be one of the four brackets.");
        }
    }

    private static boolean isOpening (char c) {
        return c == '[' || c == '{' || c == '<' || c == '(';
    }

    private static boolean isClosing (char c) {
        return c == ']' || c == '}' || c == '>' || c == ')';
    }

    public static int[] getTypes(String s) {
        return new int[]{((int) s.chars().filter(e -> e == '{' || e == '[' || e == '(' || e == '<').count()),
                ((int) s.chars().filter(e -> e == '}' || e == ']' || e == ')' || e == '>').count())};
    }

    public static Integer[] getKinds(String s) {
        Integer curly = 0, paren = 0, square = 0, angle = 0;
        for (char e : s.toCharArray()) {
            if (e == '{' || e == '}') curly++;
            if (e == '(' || e == ')') paren++;
            if (e == '[' || e == ']') square++;
            if (e == '<' || e == '>') angle++;
        }
        return new Integer[]{curly, paren, square, angle};
    }
}