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

public class RegularBracketSequence {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String seq = in.next();
		HashMap<Character, Character> myMap = new HashMap<Character, Character>();
		myMap.put('<', '>');
		myMap.put('(', ')');
		myMap.put('[', ']');
		myMap.put('{', '}');
		
		int count = 0;
		
		int open = 0;
		int close = 0;
		
		for (int i = 0; i < seq.length(); i++) {
			char curr = seq.charAt(i);
			if (curr == '<' || curr == '(' || curr == '[' || curr == '{') {
				open++;
			}
			if (curr == '>' || curr == ')' || curr == ']' || curr == '}') {
				close++;
			}
		}
		if (open != close) {
			System.out.println("Impossible");
			return;
		}
		
		
		Stack<Character> st = new Stack<Character>();
		
		for (int i = 0; i < seq.length(); i++) {
			char curr = seq.charAt(i);
			if (curr == '<' || curr == '(' || curr == '{' || curr == '[') {
				st.push(curr);
			} else if (!st.empty()) {
				if (myMap.get(st.peek()) != curr) {
					count++;
				}
				st.pop();
			} else {
				System.out.println("Impossible");
				return;
			}
		}

		System.out.println(count);
		
	}
}
