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.*;
import java.util.Stack;
import java.util.StringTokenizer;

public final class Main {
    public static void main(String[] args) throws IOException {
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String brackets = new StringTokenizer(in.readLine().trim()).nextToken();

        int result = solution(brackets);
        out.write((result == -1 ? "Impossible" : result)+"\n");
        out.flush();
    }

    private static boolean isOpening(char c)
    {
        return c == '<' || c == '(' || c == '{' || c == '[';
    }
    private static int solution(String brackets) {
        Stack<Character> stack = new Stack<>();

        int replaces = 0;
        for(char bracket: brackets.toCharArray())
        {
            if(isOpening(bracket)){
                stack.push(bracket);
            } else{
                if(stack.isEmpty()){
                    return -1;
                }

                if(!matches(stack.pop(), bracket)){
                    replaces++;
                }
            }
        }
        return stack.isEmpty() ? replaces : -1;
    }

    private static boolean matches(char left, char right) {
        return (left == '<' && right == '>') || (left == '(' && right == ')') || 
                (left == '{' && right == '}') || (left == '[' && right == ']');
    }
}