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.Scanner;
import java.util.Stack;


public class C4 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String str = in.next();
		Stack<Character> s = new Stack<Character>();
		int c = 0;
		for(int i = 0; i < str.length(); i++) {
			char ch = str.charAt(i);
			if(ch == '<' || ch == '{' || ch == '[' || ch == '(')
				s.add(ch);
			else if(s.isEmpty())
				fail();
			else {
				if(ch == ')') {
					if(s.pop() != '(')
						c++;
				}
				else if(s.pop() + 2 != ch)
					c++;
			}
		}
		if(s.isEmpty())
			System.out.println(c);
		else fail();
	}
	
	static void fail() {
		System.out.println("Impossible");
		System.exit(0);
	}
}
