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

public class ReplaceToMakeRegularBracketSequence {

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


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.next();

        Stack<Character> s = new Stack<>();

        int changes = 0;
        for(int i = 0; i < str.length(); ++i) {
            char c = str.charAt(i);
            if (s.isEmpty()) {
                if (!isOpening(c)) {
                    System.out.println("Impossible");
                    return;
                }
                else{
                    s.push(c);
                }
            }
            else {
                char top = s.peek();
                if (c == ')' && top == '(') {
                    s.pop();
                    continue;
                }
                else if (c == '}' && top == '{') {
                    s.pop();
                    continue;
                }
                else if (c == '>' && top == '<') {
                    s.pop();
                    continue;
                }
                else if (c == ']' && top == '[') {
                    s.pop();
                    continue;
                }
                else if (!isOpening(c) && isOpening(top)) {
                    changes++;
                    s.pop();
                    continue;
                }
                else if (!isOpening(c)) {
                    System.out.println("Impossible");
                    return;
                }
                else{
                    s.push(c);
                }
            }
        }

        if (s.isEmpty()) {
            System.out.println(changes);
        }
        else {
            System.out.println("Impossible");
        }
    }
}
