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 C{
	
	public static boolean left (char c){
		if(c == '[' || c == '{' || c == '(' || c =='<'){
			return true;
		}

		else {
			return false;
		}
	}

	public static char partner (char c){
		if(c == ']'){
			return '[';
		}

		else if(c=='>'){
			return '<';
		}

		else if(c==')'){
			return '(';
		}
		else if(c=='}'){
			return '{';
		}
		return ' ';
	}

	public static int fixBrackers (String s){

			int count = 0;
			Stack <Character> stack = new Stack <Character> ();

			for(int i=0; i<s.length(); i++){
				if(left(s.charAt(i))){
					stack.push(s.charAt(i));
					//System.out.println("Added" + stack.peek());
					
				}

				else{
					if(stack.empty()){
						return -1;
					}

					else if(partner(s.charAt(i))==stack.peek()){
						//System.out.println("removed" + stack.peek());
						stack.pop(); 
					}

					else{
						//System.out.println("removed/changed" + stack.peek());
						stack.pop();
						

						count++; 
					}
				}
			}



			if(stack.empty()){
				return count;
			}

			else{
				return -1; 
			}
		

	}

	public static void main (String args []){

		String a;
		Scanner in = new Scanner(System.in);

		a = in.nextLine();

		if(fixBrackers(a)==-1){
			System.out.println("Impossible");
		}

		else{
			System.out.println(fixBrackers(a));
		}


	}
}