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
//package problemset;

import java.util.Scanner;
import java.util.Stack;
import java.util.HashMap;

public class P612C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		
		Stack<Integer> st = new Stack<Integer>();
		HashMap<Character, Character> cc = new HashMap<Character, Character>();
		cc.put('[', ']');
		cc.put('<', '>');
		cc.put('{', '}');
		cc.put('(', ')');
		long ans = 0;
		for (int i = 0, l = s.length(); i < l; i++){
			char c = s.charAt(i);
			
			if (c == '[' || c == '<' || c == '{' || c=='('){
				st.push(i);
			} else{
				if (st.isEmpty()){
					System.out.println("Impossible");
					return;
				}
				if (cc.get(s.charAt(st.pop())) != s.charAt(i)){
					ans++;
				}
			}
		}
		if (!st.isEmpty()){
			System.out.println("Impossible");
		} else {
			System.out.println(ans);
		}
		
	}

}
