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.io.*;
import java.math.*;
import java.text.*;

public class Brackets {

	public static void solve(String s) {
		LinkedList<Character> stack = new LinkedList<>();
		int tot = 0;
		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			if (c == '<' || c == '{' || c == '[' || c == '(') {
				stack.push(c);
			} else {
				if (stack.isEmpty()) {
					System.out.println("Impossible");
					return;
				}
				char r = stack.pop();
				if (
					(r == '<' && c != '>') ||
					(r == '{' && c != '}') ||
					(r == '[' && c != ']') ||
					(r == '(' && c != ')')
				  )
					tot++;
			}
		}
		if (stack.isEmpty()) {
			System.out.println(tot);
		} else {
			System.out.println("Impossible");
		}
	}

	public static void main(String[] args) {
		try {
			InputStreamReader is = new InputStreamReader(System.in);
			BufferedReader br = new BufferedReader(is);
			solve(br.readLine() );
		} catch (Exception e) {e.printStackTrace();}
	}

}

