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.*;
import java.util.Scanner;
import java.io.*;
import javax.lang.model.util.ElementScanner6; 
import static java.lang.System.out;
import java.util.Stack;

public class C612
{

   
    
    public static void main(String args[])
    {  

        FastReader in=new FastReader();
        PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int tc=1;
     
        //tc=in.nextInt();
        while(tc-->0)
        {
            
        	char ch[]=in.nextLine().toCharArray();
        	
        	Stack<Character> st=new Stack<>();
        	int ans=0;
        	int f=1;
        	HashMap<Character,Character> hm=new HashMap<>();
        	hm.put('{','}');
        	hm.put('(',')');
        	hm.put('[',']');
        	hm.put('<','>');
        	
        	for(int i=0;i<ch.length;i++)
        	{
				if(ch[i]=='('||ch[i]=='{'||ch[i]=='<'||ch[i]=='[')
				{
					st.add(ch[i]);
				}
				else
				{
					//pr.println("peek "+st.peek()+" imput "+ch[i]);
					if(st.isEmpty())
					{
						f=0;
						break;
					}
					else
					{
					
						if(hm.get(st.peek())==ch[i])
						{
							st.pop();
							
						}
						else
						{	
							st.pop();
							
							 ans++;
						 }
					}
				}
			}
			if(f==1&&st.isEmpty())pr.println(ans);
			
			else pr.println("Impossible");
        	

        }
        pr.flush();
    }


    

    static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
    
    
	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
    }

    
    static class FastReader
    { 
        BufferedReader br; 
        StringTokenizer st; 

        public FastReader() 
        { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 

        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 

        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 

        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 

        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 

        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
}
