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.InputStreamReader;
import java.util.HashSet;
import java.util.Set;
import java.util.Stack;

public class Main {
    static BufferedReader jin = new BufferedReader( new InputStreamReader( System.in ) );
    static Stack<Character> stack = new Stack<>();
    static Set<Character> open_ch = new HashSet<>();
    static Set<Character> clos_ch = new HashSet<>();

    public static void main(String[] args) throws Exception {
        open_ch.add('<'); clos_ch.add('>');
        open_ch.add('{'); clos_ch.add('}');
        open_ch.add('['); clos_ch.add(']');
        open_ch.add('('); clos_ch.add(')');
        int n = 0;
        char[] line = jin.readLine().toCharArray();

        for(char ch : line) {
            if( open_ch.contains(ch) ) stack.push(ch);
            else if( !stack.empty() ) {
                char head = stack.pop();
                if( clos_ch.contains(head) ) { System.out.print("Impossible"); return; }
                else if( !ifGetClosing(head, ch) ) n++;
            } else { System.out.print("Impossible"); return; }
        }

        System.out.print( (stack.empty() && n >= 0)? n : "Impossible");
    }

    static boolean ifGetClosing(char key, char clos) {
        switch( key ) {
            case '<' : return ( '>' == clos )? true : false;
            case '{' : return ( '}' == clos )? true : false;
            case '[' : return ( ']' == clos )? true : false;
            case '(' : return ( ')' == clos )? true : false;
        }
        return false;
    }
}