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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;

public class C {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		String line = sc.next();
		boolean possible = true;
		int open = 0, ans = 0;
		Stack<Integer> opened = new Stack<Integer>();
		for(int i = 0; possible && i < line.length(); ++i)
		{
			char c = line.charAt(i);
			if(isOpening(c))
				++open;
			else
				--open;
			if(open < 0)
				possible = false;
			else
			{
				
				if(isOpening(c))
				{
					int t = getTypeOpen(c);
					opened.push(t);
				}
				else
				{
					int t = getTypeClosed(c);
					if(opened.peek() != t)
						++ans;
					opened.pop();
				}
			}
		}
		if(open == 0 && possible)
		{
			
			
			out.println(ans);
		}
		else
			out.println("Impossible");
		out.flush();
		out.close();
	}
	
	static boolean isOpening(char c)
	{
		return c == '<' || c == '(' || c == '{' || c == '[';
	}
	
	static int getTypeOpen(char c)
	{
		switch(c)
		{
		case '<': return 0;
		case '{': return 1;
		case '[': return 2;
			default: return 3;
		}
	}
	
	static int getTypeClosed(char c)
	{
		switch(c)
		{
		case '>': return 0;
		case '}': return 1;
		case ']': return 2;
			default: return 3;
		}
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(FileReader fileReader) throws FileNotFoundException{	br = new BufferedReader(fileReader);}

		public Scanner(InputStream s) throws FileNotFoundException{	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}


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

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

		public String nextLine() throws IOException {return br.readLine();}

		public boolean ready() throws IOException {return br.ready(); }


	}

}