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;

/**
 * http://codeforces.com/contest/612/problem/C
 * 
 * @author Callie
 *
 */

public class ReplaceBracketSequence {

	public static int solve(String str) {
		if (str.length() % 2 != 0) // this is checked below, but including it
									// here may be able to speed up performance
									// time without having to check the entire
									// string
			return -1;

		int res = 0;
		Map<Character, Character> map = new HashMap<Character, Character>();
		map.put('(', ')');
		map.put('<', '>');
		map.put('{', '}');
		map.put('[', ']');

		Stack<Character> stack = new Stack<Character>();

		for (int i = 0; i < str.length(); i++) {
			char c = str.charAt(i);
			if (map.containsKey(c))
				stack.push(c);
			else { // assumes closing parentheses
				if (stack.isEmpty()) {
					return -1;
				}
				char top = stack.peek();
				if (map.get(top) != c) {
					res++;
				}
				stack.pop();
			}
		}
		if(stack.isEmpty())
			return res;
		else
			return -1;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		String str = in.nextLine();
		
		int res = solve(str);

		if (res == -1) {
			System.out.println("Impossible");
		} else {
			System.out.println(res);
		}
	}
}