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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Stack;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author safadurimo
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            String s = in.nextLine();
            Stack<Character> st = new Stack<Character>();

            int x = 0;
            for (int i = 0; i < s.length(); i++) {
                char c = s.charAt(i);
                if (c == '}' || c == ')' || c == '>' || c == ']') {
                    if (st.isEmpty()) {
                        x = -1;
                        break;
                    } else {
                        char d = st.pop();
                        if (d != trans(c)) x++;
                    }
                } else st.push(c);

            }
            if (x < 0 || !st.isEmpty()) {
                out.println("Impossible");
            } else {
                out.println(x);
            }
        }

        public static char trans(char c) {
            switch (c) {
                case '}':
                    return '{';
                case ')':
                    return '(';
                case '>':
                    return '<';
                case ']':
                    return '[';

            }
            return 'a';
        }

    }
}