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

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String str = "";
		if (in.hasNext())
			str = in.next();
		in.close();
		if (str.length() % 2 != 0) {// 奇数是不可能对的
			System.out.println("Impossible");
			return;
		} else {
			int num = 0;
			char[] tmp = str.toCharArray();
			Stack<Character> stack = new Stack<>();
			for (int i = 0; i < str.length(); i++) {
				char curr = str.charAt(i);
				// 栈为空，来了右括号
				if (stack.size() == 0 && (curr == '}' || curr == ']' || curr == ')' || curr == '>')) {
					System.out.println("Impossible");
					return;
				}
				// 遇到左括号就添加
				if (curr == '{' || curr == '[' || curr == '(' || curr == '<') {
					stack.push(curr);
					// 非左括号非右括号的跳过
				}  else {// 遇到右括号就匹配后弹出
					if ((curr == '}' && stack.peek() != '{') || (curr == ']' && stack.peek() != '[')
							|| (curr == ')' && stack.peek() != '(') || (curr == '>' && stack.peek() != '<')) {
						num++;
					}
					stack.pop();
				}
			}
			if(stack.size()!=0) {
				System.out.println("Impossible");
				return;
			}
			System.out.println(num);
		}
	}
}