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.*;
import java.io.*;

public class RBS {
    
     
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    
    public static void main(String[] args) throws IOException{
        
        var seq = reader.readLine().trim().toCharArray();

        if(seq.length % 2 != 0){

            System.out.println();

        }

        var stack = new Stack<Character>();


        int a = 0;

        var o_c = new HashMap<Character, Character>(4);
        o_c.put('<', '>');
        o_c.put('[', ']');
        o_c.put('{', '}');
        o_c.put('(', ')');

        var c_o = new HashMap<Character, Character>(4);
        c_o.put('>', '<');
        c_o.put(']', '[');
        c_o.put('}', '{');
        c_o.put(')', '(');


        for(char c: seq){

            if(o_c.containsKey(c))
                stack.push(c);

            else {

                if(stack.isEmpty()){

                    a = -1;
                    break;
                }

                char open = stack.pop();
                a = c_o.get(c) == open ? a : a + 1;
            }
        }

        if(stack.size() > 0)
          a = -1;

        int b = 0, n = seq.length; stack.clear();
        for(int i = n - 1; i >= 0; --i){
            
            char c = seq[i];

            if(c_o.containsKey(c))
                stack.push(c);

            else {

            if(stack.isEmpty()){

                b = -1;
                break;
            }

                char close = stack.pop();
                b = o_c.get(c) == close ? b : b + 1;
            }
        }

        if(stack.size() > 0)
          b = -1;

        if(a == - 1 && b == -1)
            System.out.println("Impossible");

        else if(a == -1)
            System.out.println(b);

        
        else if(b == -1)
            System.out.println(a);

        else
            System.out.println(Math.min(a, b));

    }


}