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 Main 
{

	/**
	 * @param args
	 */
	public static void main(String[] args) 
	{
		// TODO Auto-generated method stub
		Scanner sc=new Scanner (System.in);
		
		Stack<Character> stack=new Stack<Character>();
		String s=sc.next();
		char c;
		boolean sign=true;
		int count=0;
		for(int i=0;i<s.length();i++)
		{
			c=s.charAt(i);
			if(c=='<')
				stack.add('>');
			else
				if(c=='{')
					stack.add('}');
				else
					if(c=='[')
						stack.add(']');
					else
						if(c=='(')
							stack.add(')');
						else
						{
							if(stack.isEmpty())
							{
								sign=false;
								break;
							}
							else
							{
								char k=stack.pop();
								if(k!=c)
									count++;	
							}
						}
		}
		if(sign&&stack.isEmpty())
			System.out.println(count);
		else
			System.out.println("Impossible");
	}

}
	   		 	 	  			 	 			 				 	  	