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.*;

public class Solution {
	
    public static void main(String[] args) {
    	Scanner scan = new Scanner(System.in);
    	
    	HashMap<Character, Integer> map = new HashMap<Character, Integer>();
    	map.put('{', 0);
    	map.put('}', 0);
    	map.put('[', 1);
    	map.put(']', 1);
    	map.put('(', 2);
    	map.put(')', 2);
    	map.put('<', 3);
    	map.put('>', 3);
    	
    	String str = scan.next();
    	Stack<Character> brackets = new Stack<Character>();
    	
    	boolean flag = true;
    	int count = 0;
    	for(int i = 0; i < str.length(); i++){
    		if(isOpening(str.charAt(i))){
    			brackets.push(str.charAt(i));
    		}
    		else{
    			if(brackets.isEmpty()){
    				flag = false;
    				break;
    			}
    			char br = brackets.pop();
    			if(! (map.get(br) == map.get(str.charAt(i)))){
    				count++;
    			}
    		}
    	}
    	if(!brackets.isEmpty()){
    		flag = false;
    	}
    	
    	
    	System.out.println(flag? count : "Impossible");
    	
    	scan.close();
    }
    
    public static boolean isOpening(char ch){
    	if(ch == '{' || ch =='[' || ch == '(' || ch == '<')
    		return true;
    	return false;
    }
}

