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 main{
	public static void main(String args[]){
		Scanner sc=new Scanner(System.in);
		
		String s1=sc.next();
		int count=0;
		Stack<Character> s=new Stack<Character>();
		boolean b=true;

		for(int i=0;i<s1.length();i++){
			if(s1.charAt(i)=='<'){	
				s.push(s1.charAt(i));	
			}
			else if(s1.charAt(i) == '>'){
				if(s.size()==0){
					b=false;
					break;
				}
				if(s.peek()!='<'){
					count++;
				}
				s.pop();
			}
			else if(s1.charAt(i) == '{'){
				s.push(s1.charAt(i));
			}
			else if(s1.charAt(i) =='}'){
				if(s.size()==0){
					b=false;
					break;
				}
				if(s.peek()!='{'){
					count++;
				}
				s.pop();
			}
			else if(s1.charAt(i) == '['){
				s.push(s1.charAt(i));
			}
			else if(s1.charAt(i) ==']'){
				if(s.size()==0){
					b=false;
					break;
				}
				if(s.peek()!='['){
					count++;
				}
				s.pop();
			}
			else if(s1.charAt(i) == '('){
				s.push(s1.charAt(i));
			}
			else if(s1.charAt(i) ==')'){
				if(s.size()==0){
					b=false;
					break;
				}
				if(s.peek()!='('){
					count++;
				}
				s.pop();
			}
		}
		
		if(!s.empty() || !b){
			System.out.print("Impossible");
		}else{
			System.out.print(count);
		}
		
		
		
	}
}