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.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

/**
 * Dec 25, 2015 | 9:09:56 PM
 * <pre>
 * <u>Description</u>
 *
 * </pre>
 *
 * @author Essiennta Emmanuel (colourfulemmanuel@gmail.com)
 */
public class ProblemC{
    
    int solve(char[] s){
        Deque<Character> stack = new ArrayDeque<>(s.length);
        int cnt = 0;
        for (int i = 0; i < s.length; i++){
            if (s[i] == '<' || s[i] == '{' || s[i] == '[' || s[i] == '(')
                stack.push(s[i]);
            else if (stack.isEmpty())
                return -1;
            else if (
                    (s[i] == '>' && stack.pop() != '<'
                    || s[i] == '}' && stack.pop() != '{'
                    || s[i] == ']' && stack.pop() != '['
                    || s[i] == ')' && stack.pop() != '('
                    ))
                cnt++;
        }
        return !stack.isEmpty() ? -1 : cnt;
    }
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        char[] s = sc.next().toCharArray();
        final int ans = new ProblemC().solve(s);
        System.out.println(ans == -1 ? "Impossible" : ans);
        sc.close();
    }
}
