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.io.*;
import java.util.*;

public class ReplaceRegBra 
{
	public static void main(String args[])throws IOException
	{
		BufferedReader br = new BufferedReader (new InputStreamReader (System.in));
		PrintWriter out = new PrintWriter (System.out);
		String s = br.readLine();
		char [] brackets = s.toCharArray();
		
		Stack <Character> t = new Stack<Character>();
		int c=0;
		
		for (int i = 0; i<brackets.length; i++)
		{
			char in = brackets[i];
			if (in == '{' || in == '[' || in == '<' || in == '(' )
				t.add(in);
			else
			{
				if (t.size() == 0)
				{
					System.out.println("Impossible");
					return;
				}
				else
				{
					if (in == '}')
					{
						if (t.peek() == '{')
							t.pop();
						else
							{t.pop(); c++;}
					}
					if (in == ']')
					{
						if (t.peek() == '[')
							t.pop();
						else
							{t.pop(); c++;}
					}
					if (in =='>')
					{
						if (t.peek() == '<')
							t.pop();
						else
							{t.pop(); c++;}
					}
					if (in == ')')
					{
						if (t.peek() == '(')
							t.pop();
						else
							{t.pop(); c++;}
					}
				}

			}
		}
		if (t.size() > 0)
			{out.println("Impossible");out.close();return;}
		out.println(c);
		out.close();
		
	}

}