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.io.*;
import java.util.Stack;
public class ReplaceToMakeRegularBracketSequence {

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader (new InputStreamReader(System.in));
        String s = br.readLine();
        int right = 0;
        int left = 0;
        Stack<Character> st = new Stack<Character>();
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(st.isEmpty() && (c == '>' || c == ']' || c == '}' || c == ')')){
                right = 0;
                left = 1;
                break;
            }
            if(!st.isEmpty() && ((st.peek() == '[' && c == ']') || (st.peek() == '(' && c == ')')
                    || (st.peek() == '<' && c == '>')|| (st.peek() == '{' && c == '}'))){
                    right--;
                    st.pop();
            }else if(!st.isEmpty() && (st.peek() == '[' || st.peek() == '{' 
                    || st.peek() == '<' || st.peek() == '(') 
                    && (c == '>' || c == ']' || c == '}' || c == ')')){
                left++;
                st.pop();
            }
            else {
            if(c == '[' || c == '<' || c == '{' || c == '(')
                right++;
            else
                left++;
            st.push(c);
            }
            
        }
         if(right!=left){
            System.out.println("Impossible");
        }else{
            System.out.println(right);
        }

    }

}
