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.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.Stack;


public class C {
	
	static void solve(String s) {
		String imp = "Impossible";
		if(s.length() % 2 == 1) {
			System.out.println(imp);
			return;
		}
		
		Stack<Character> open = new Stack<Character>();
		
		Set<Character> o = new HashSet<Character>();
		o.add('[');
		o.add('<');
		o.add('{');
		o.add('(');
		
		Map<Character, Character> hm = new HashMap<Character, Character>();
		hm.put(']', '[');
		hm.put('>', '<');
		hm.put('}', '{');
		hm.put(')', '(');
		
		if(o.contains(s.charAt(0))) {
			open.push(s.charAt(0));
		}
		else {
			System.out.println(imp);
			return;
		}
		long cnt = 0;
		
		for(int i = 1; i < s.length(); i++) {
			char c = s.charAt(i);
			if(o.contains(c)) {
				open.push(c);
			}
			else if(hm.containsKey(c)) {
				if(open.isEmpty()) {
					System.out.println(imp);
					return;
				}
				if(open.pop() != hm.get(c)) {
					cnt++;
				}
			}
		}
		if(open.isEmpty()) {
			System.out.println(cnt);
		}
		else {
			System.out.println(imp);
		}
	}

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s = br.readLine();
		br.close();
		
		solve(s);
		
	}

}