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.Scanner;
import java.util.Stack;

public class Main {
    public Main() {
        super();
    }

    public int validRBS(String s) {
        int ret=0;

        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char theChar = s.charAt(i);
            //System.out.println(theChar);
            if(stack.empty() && (theChar == ')' || theChar == '}' || theChar == ']' || theChar == '>') )
                return -1;
            
            if (theChar == '(' || theChar == '{' || theChar == '[' || theChar == '<')
                stack.push(theChar);
            else 
            if(!(theChar == ')' || theChar == '}' || theChar == ']' || theChar == '>'))
              continue;
            else
            {
                if (stack.empty()) // 栈中已无左括号，此时字符为右括号，无法匹配。
                    return -1;
                char preChar = stack.peek();
                if ((preChar == '{' && theChar != '}') || (preChar == '(' && theChar != ')') || (preChar == '[' && theChar != ']') || (preChar == '<' && theChar != '>')){
                    ret++;
                }
                stack.pop();
            }
        }
        //System.out.println(stack.empty());
        if(!stack.empty())return -1;
        
        return ret;
        
    }
    public static void main(String[] args) {
        Main m = new Main();

        Scanner scan = new Scanner(System.in);
        //System.out.println("next方式接收：");
        String s="]]";
        if (scan.hasNext()) s=scan.next();
        scan.close();
        int r=0;
        r=m.validRBS(s);
        
        if(r<0)System.out.println("Impossible");
        else
        System.out.println(r);
    }
}