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 ER4C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn = new Scanner(System.in);
		String s = scn.next();
		Stack<Integer> stack = new Stack<>();
		// ( 1
		// { 2
		// [ 3
		int count = 0;
		int i = 0;
		while (i < s.length()) {
			char c = s.charAt(i);
			if (c == '(') {
				stack.push(1);
			} else if (c == '{') {
				stack.push(2);
			} else if (c == '[') {
				stack.push(3);
			} else if (c == '<') {
				stack.push(4);
			} else if (c == '>') {
				if (stack.size() == 0) {
					System.out.println("Impossible");
					return;
				}
				Integer p = stack.peek();

				if (p == 4) {
					stack.pop();
				} else {
					stack.pop();
					count++;
				}
			} else if (c == ')') {
				if (stack.size() == 0) {
					System.out.println("Impossible");
					return;
				}
				Integer p = stack.peek();

				if (p == 1) {
					stack.pop();

				} else {
					stack.pop();
					count++;
				}
			} else if (c == ']') {
				if (stack.size() == 0) {
					System.out.println("Impossible");
					return;
				}
				Integer p = stack.peek();
				if (p == 3) {
					stack.pop();
				} else {
					stack.pop();
					count++;
				}
			} else if (c == '}') {
				if (stack.size() == 0) {
					System.out.println("Impossible");
					return;
				}
				Integer p = stack.peek();

				if (p == 2) {
					stack.pop();
				} else {
					stack.pop();
					count++;
				}
			}
			i++;
		}
		if (stack.size() > 0) {
			System.out.println("Impossible");
			return;
		}
		System.out.println(count);
	}

}