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.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Stack;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		Stack<Character> stack = new Stack<Character>();
		Map<Character,Boolean> open = new HashMap<Character,Boolean>();
		open.put('{', true);
		open.put('(', true);
		open.put('<', true);
		open.put('[', true);
		open.put('}', false);
		open.put('>', false);
		open.put(']', false);
		open.put(')', false);
		Map<Character,Character> match = new HashMap<Character,Character>();
		match.put('}', '{');
		match.put('>', '<');
		match.put(']', '[');
		match.put(')', '(');
		boolean possible = true;
		int count = 0;
		for(int i=0;i<s.length();i++){
			if(open.get(s.charAt(i))){
				stack.add(s.charAt(i));
			}
			else{
				if(stack.size()==0){
					possible = false;
					break;
				}
				else{
					char c = stack.pop();
					if(c != (char)match.get(s.charAt(i))){
						count++;
					}
				}
			}
		}
		if(possible && stack.size()==0){
			System.out.println(count);
		}
		else{
			System.out.println("Impossible");
		}
	}
}