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.util.Scanner;
import java.util.Stack;

public class C {
	public static void main(String[] args) {
		try(Scanner scan = new Scanner(System.in)) {
			String s = scan.nextLine();
			String set = "(){}[]<>";
			
			int sum = 0;
			boolean valid=true;
			for(int i=0;i<s.length();i++){
				if(set.indexOf(s.charAt(i))%2==0)sum++;
				else sum--;
				
				if(sum<0){valid=false;break;}
			}
			if(!valid||sum!=0){System.out.println("Impossible");return;}
			
			Stack<Character> stack = new Stack<>();
			int ans = 0;
			for(int i=0;i<s.length();i++){
				int id = set.indexOf(s.charAt(i));
				if(id%2==0)stack.add(s.charAt(i));
				else if(set.indexOf(stack.peek())==id-1){
					stack.pop();
				}
				else{
					ans++;
					stack.pop();
				}
			}
			System.out.println(ans);
		}
	}
}
