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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class C {

    static StringTokenizer st;
    static BufferedReader br;
    static PrintWriter pw;
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        String s = next();
        char[]st = new char[s.length()+1];
        int n = 0, ans = 0;
        for (int i = 0; i < s.length(); i++) {
            char w = s.charAt(i);
            if (open(w))
                st[++n] = w;
            else {
                if (n==0) {
                    System.out.println("Impossible");
                    return;
                }
                if (st[n] != opposite(w))
                    ans++;
                n--;
            }
        }
        if (n != 0)
            System.out.println("Impossible");
        else
            System.out.println(ans);
        pw.close();
    }
    private static char opposite(char w) {
        if (w==')')
            return '(';
        if (w=='>')
            return '<';
        if (w=='}')
            return '{';
        return '[';
    }
    private static boolean open(char w) {
        return w=='<' || w=='(' || w=='[' || w=='{';
    }
    private static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    private static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    private static String next() throws IOException {
        while (st==null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }
}