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
import java.util.Stack;
import java.util.Scanner;

public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        char[] s = in.nextLine().toCharArray();
        Stack<Character> stack = new Stack<>();
        int count = 0;

        char[] map = new char[256];
        map['['] = ']';
        map['<'] = '>';
        map['{'] = '}';
        map['('] = ')';

        for (char c : s) {
            if (c == '{' || c == '(' || c == '<' || c == '[') {
                stack.push(c);
            } else if (stack.isEmpty()) {
                count = Integer.MIN_VALUE;
            } else {
                char top  = stack.pop();
                if (map[top] != c) count += 1;
            }
        }

        if (count < 0 || !stack.isEmpty()) {
        System.out.println("Impossible");
        } else {
            System.out.println(count);
        }
    }
}
