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.Scanner;
import java.util.Stack;

public class BracketSequence {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		char[] arr = sc.next().toCharArray();
		int len = arr.length;
		
		Stack<Character> stack = new Stack<Character>();
		int count = 0;
		int left = 0; int right = 0;
		
		for(int i=0; i<len; i++){
			if(isOpening(arr[i])) {
				stack.add(arr[i]);
				left++;
			} else {
				right++;
				if(stack.isEmpty()) {
					System.out.println("Impossible");
					System.exit(0);
				}
				char tempChar = stack.pop();
				if(isSameType(tempChar, arr[i])) continue;
				count++;
			}
		}		
		
		if(left != right) {
			System.out.println("Impossible");
			System.exit(0);
		}
		
		System.out.println(count);
		
	}
	
	public static boolean isOpening(char c) {
		return (c == '<' || c == '{' || c=='[' || c=='(');
	}
	
	static boolean isSameType(char a, char b) {
		if(a == '(' && b == ')') return true;
		if(a == '{' && b == '}') return true;
		if(a == '[' && b == ']') return true;
		if(a == '<' && b == '>') return true;
		return false;
	}
}
