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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Stack;

public class C {

	public static void main(String[] args) throws IOException {
		   BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
		    String in = input.readLine();
		    char[] init = in.toCharArray();
		    
		    int  n = init.length;
		    int tl = 0;
		    int tr = 0;
		    
		    int[] count = new int[Character.MAX_VALUE + 1];
		    // <>
		    int al = 0;
		    int ar = 0;
		    
		    // {}
		    int bl = 0;
		    int br = 0;
		    
		    // []
		    int cl = 0;
		    int cr = 0;
		    
		    // ()
		    int dl = 0;
		    int dr = 0;

		    Stack<Character> open = new Stack<Character>();
		    int r = 0;
		    for(int i = 0; i < n; i++){
		    	if(init[i] == '<' || init[i] == '[' || init[i] == '{' || init[i] == '('){
		    		open.push(init[i]);
		    	} else {
		    		if(open.isEmpty()){
				    	System.out.println("Impossible");
				    	return;
		    		}
		    		
		    		char c = open.pop();
		    		
		    		if(c == '<'){
		    			if(init[i] != '>'){
		    				r++;
		    			}
		    		}
		    		
		    		if(c == '{'){
		    			if(init[i] != '}'){
		    				r++;
		    			}
		    		}
		    		
		    		if(c == '['){
		    			if(init[i] != ']'){
		    				r++;
		    			}
		    		}
		    		
		    		if(c == '('){
		    			if(init[i] != ')'){
		    				r++;
		    			}
		    		}
		    	}
		    }
		    
		    if(!open.isEmpty()){
		    	System.out.println("Impossible");
		    	return;
		    }
		    
		    System.out.println(r);

	


	}
	
	
}