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.*;

public class EbuchieSkobki {

    public static void main(String[] args) {
       new EbuchieSkobki().solve();
    }

    private void solve() {
        final Scanner in = new Scanner(System.in);
        Stack<Character> sc = new Stack<>();
        Map<Character, Character> hm = new HashMap<>();
        Set<Character> st = new HashSet<>();
        st.add('[');
        st.add('(');
        st.add('{');
        st.add('<');
        hm.put('}', '{');
        hm.put(']', '[');
        hm.put('>', '<');
        hm.put(')', '(');

        int count = 0;
        final String str = in.nextLine();

        for (int i = 0; i < str.length(); i++) {
            if (sc.isEmpty() & hm.containsKey(str.charAt(i))) {
                System.out.println("Impossible");
                return;
            }

            if (st.contains(str.charAt(i))) {
                sc.push(str.charAt(i));
                continue;
            }

            if (sc.peek() == hm.get(str.charAt(i))) {
                sc.pop();
            } else {
                sc.pop();
                count++;
            }

            if ((i == str.length() - 1) & !sc.isEmpty()) {
                System.out.println("Impossible");
                return;
            }
        }

        if (!sc.isEmpty()) {
            System.out.println("Impossible");
            return;
        }
        
        System.out.println(count);
    }
}