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.Map;
import java.util.Scanner;
import java.util.Stack;

public class C {

    static Map<Character, Character> match = new HashMap<Character, Character>() {{
        put('(', ')');
        put(')', '(');
        put('<', '>');
        put('>', '<');
        put('{', '}');
        put('}', '{');
        put(']', '[');
        put('[', ']');
    }};

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        char[] x = s.toCharArray();
        Stack<Character> stack = new Stack<>();
        int result = 0;
        for (char c : x) {
            boolean isOpening = isOpening(c);
            if (stack.isEmpty()) {
                if (isOpening) {
                    stack.push(c);
                } else {
                    System.out.println("Impossible");
                    return;
                }
            } else {
                char matching = match.get(c);
                char last = stack.peek();
                if (!isOpening) {
                    if (last != matching) {
                        result++;
                    }
                    stack.pop();
                } else {
                    stack.push(c);
                }
            }
        }
        if (stack.isEmpty()) {
            System.out.println(result);
        } else {
            System.out.println("Impossible");
        }
        sc.close();
    }

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