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 ReplaceRegularStrings{
	public static void main(String []args) throws Exception{
		
		java.io.InputStreamReader isr=new java.io.InputStreamReader(System.in);
		java.io.BufferedReader br=new java.io.BufferedReader(isr);
		
		String str=br.readLine();
		Stack<Character> stack=new Stack<Character>();
		int flag=0,count=0;
		for(int i=1;i<=str.length();i++){
			if(str.charAt(i-1)=='<' || str.charAt(i-1)=='[' || str.charAt(i-1)=='{' || str.charAt(i-1)=='('){
				stack.push(str.charAt(i-1));
			}else if(str.charAt(i-1)=='>' || str.charAt(i-1)==']' || str.charAt(i-1)=='}' || str.charAt(i-1)==')'){
				if(stack.size()>0){
					char ch=stack.pop();
					if(ch=='<' && (str.charAt(i-1)==']' || str.charAt(i-1)=='}' || str.charAt(i-1)==')')){
						count++;
					}else if(ch=='[' && (str.charAt(i-1)=='>' || str.charAt(i-1)=='}' || str.charAt(i-1)==')')){
						count++;
					}else if(ch=='{' && (str.charAt(i-1)==']' || str.charAt(i-1)=='>' || str.charAt(i-1)==')')){
						count++;
					}else if(ch=='(' && (str.charAt(i-1)==']' || str.charAt(i-1)=='}' || str.charAt(i-1)=='>')){
						count++;
					}
				}else{
					flag=1;
					break;
				}
			}
		}
		if(stack.size()>0){
			flag=1;
		}
		if(flag==1){
			System.out.println("Impossible");
		}else{
			System.out.println(count);
		}
		
		
	}
}