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 lab612 {
    public static void main (String args[]) {
        Scanner sc = new Scanner(System.in);
        char [] ar = sc.next().toCharArray();
        Stack<Character> st = new Stack<>();
        char[] op = new char[]{'<', '{', '[', '('};
        char[] cl= new char[]{'>', '}', ']', ')'};
        int f = 0;
        int ans = 0;
        for(int i = 0; i<ar.length; i++)
        {
            Character ch = ar[i];
            int j=0;int k=0;
            for(j = 0; j<4; j++)
                if(ch==op[j]) break;
            for(k = 0; k<4; k++)
                if(ch==cl[k]) break;
            if(j<4){
                st.push(ch);
            }
            else{
                if(st.empty()){
                    f = 1;
                    break;
                }
                else{
                    char top = st.peek();
                    if(st.peek()!=op[k]) ans++;
                    st.pop();
                }
            }
        }
        if(f==1 || st.empty()==false) System.out.println("Impossible");
        else System.out.println(ans);
    }
}