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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class ReplaceRBS {

	public static void main(String[] args){
		// Use BufferedReader rather than RandomAccessFile; it's much faster
	    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
	                                                  // input file name goes above
	    PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	    // Use StringTokenizer vs. readLine/split -- lots faster
	    StringTokenizer st = null;
	   
							  // Get line, break into tokens
	    
	   //enter program here
	    long startTime = System.currentTimeMillis();
	    int c2 = 0;
	    int counter = 0;
	    Stack<String> RBS = new Stack<String>();
	    boolean error = false;
	    search:
	    try {
			while((c2 = f.read()) != -1) {
			    char c = (char) c2;
			    //System.out.println(character);
			    if (c =='[' || c == '(' || c=='<' || c=='{'){
			    	RBS.add(c+"");
			    }else if (c==']' || c==')' || c=='>' || c=='}'){
			    	String curr = RBS.pop();
			    	if (curr.charAt(0)=='[' && c!=']' || curr.charAt(0)=='(' && c!=')' || curr.charAt(0)=='<' && c!='>' || curr.charAt(0)=='{' && c!='}')
			    		counter +=1;
			    }else{
			    	break search;
			    }
			    
			}
		} catch (Exception e) {
			// TODO Auto-generated catch block
			error = true;
		}       
	    
	    if (RBS.size()==0 && !error)
	    System.out.println(counter);
	    else
	    System.out.println("Impossible");
	    //System.out.println(System.currentTimeMillis()-startTime);
	    // answer output

    	out.close();	    
	    System.exit(0);                     
	}
}