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.Scanner;
import java.util.Stack;

public class C {

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

        String bracket = input.next();

        int start = 1, startIndex = 0, counter = 0;
        if (check(bracket.charAt(0)) == -1) System.out.println("Impossible");
        else {
            Stack <Character> q = new Stack<Character>();
            q.push(bracket.charAt(0));
            for (int i = 1; i < bracket.length(); i++) {
                start = start + check(bracket.charAt(i));
                if (check(bracket.charAt(i)) == 1) q.push(bracket.charAt(i));
                else if (!q.empty()){
                    Character b = q.pop();
                    if (b == '(' && bracket.charAt(i) != ')') counter++;
                    if (b == '<' && bracket.charAt(i) != '>') counter++;
                    if (b == '{' && bracket.charAt(i) != '}') counter++;
                    if (b == '[' && bracket.charAt(i) != ']') counter++;
                }
                if (start < 0) break;
            }
            if (start != 0) System.out.println("Impossible");
                else System.out.println(counter);
        }
    }
    static public int check(char ch){
        if  (ch == '(' || ch == '<' || ch == '{' || ch == '[') return 1;
        return -1;
    }
}