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.*;
public class Main {
public static void main(String [] args)throws Exception{
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	String s = br.readLine();
	Stack<Character>st = new Stack<Character>();
	Character array[] = {' ','<','>','{','}','[',']','(',')'};
	int ans = 0;
	for(int i = 0;i < s.length();i++){
		if(s.charAt(i) == '<' || s.charAt(i) == '(' || s.charAt(i) == '{' || s.charAt(i) == '[')
			st.push(s.charAt(i));
		else{
			if(st.isEmpty()){
				System.out.print("Impossible");
				return;
			}
			char A = st.pop();
			boolean flag = false;

			for(int j = 1;j <= 7;j++)
				if(array[j] == A && array[j + 1] == s.charAt(i))
				flag = true;

			if(!flag)
			ans++;
		}
	}
	if(!st.isEmpty())
		System.out.println("Impossible");
	else System.out.println(ans);
}
}