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:
//package edu4;

import java.util.Scanner;

public class p3 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = new String();
		s = sc.nextLine();
		int base = 0;
		for (int i = 0; i < s.length(); i++) {
			switch (s.charAt(i)) {
			case '{':
				base++;
				break;
			case '(':
				base++;
				break;
			case '<':
				base++;
				break;
			case '[':
				base++;
				break;
			default:
				base--;
			}
			if(base<0){
				System.out.println("Impossible");
				return;
			}
		}
		if(base!=0){
			System.out.println("Impossible");
			return;
		}
		int[] stack=new int[s.length()+1];
		int point=-1;
		int ans=0;
		for(int i=0;i<s.length();i++){
			switch (s.charAt(i)) {
			case '{':
				stack[++point]=i;
				break;
			case '(':
				stack[++point]=i;
				break;
			case '<':
				stack[++point]=i;
				break;
			case '[':
				stack[++point]=i;
				break;
			default:
				// System.out.println(s.charAt(i)+" "+s.charAt(stack[point]));
				// System.out.println((int)s.charAt(i)+"
				// "+(int)s.charAt(stack[point]));
				// System.out.println();
				if(Math.abs((int)s.charAt(i)-(int)s.charAt(stack[point]))<=2)
					point--;
				else{
					ans++;
					point--;
				}
			}
		}
		System.out.println(ans);
	}

}