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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		String str = scan.next();
		char[] parens = str.toCharArray();
		int n = parens.length;
		String Z = "<{([";
		int bal = 0;
		boolean possible = true;
		for(int i = 0; i < n; i++) {
			if(Z.contains(parens[i]+"")) bal++;
			else bal--;
			if(bal < 0) possible = false;
		}
		if(bal != 0) possible = false;
		if(!possible) System.out.println("Impossible");
		if(possible) {
			int[] A = new int[n/2];
			int[] B = new int[n/2];
			int a = 0;
			ArrayList<Integer> Q = new ArrayList<Integer>();
			for(int i = 0; i < n; i++) {
				if(Z.contains(parens[i]+"")) {
					A[a] = i;
					Q.add(a);
					a++;
				}
				else {
					B[Q.get(Q.size()-1)] = i;
					Q.remove(Q.size()-1);
				}
			}
			int ans = 0;
			for(int i = 0; i < n/2; i++) {
				if(!pair(parens[A[i]], parens[B[i]])) ans++;
			}
			System.out.println(ans);
		}
	}
	
	public static boolean pair(char x, char y) {
		String a = "";
		a += x;
		a += y;
		return a.equals("<>") || a.equals("()") || a.equals("[]") || a.equals("{}");
	}
}