import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Solution to Stones of Yin & Yang
 * 
 * @author vanb
 */
public class yinyang_vanb
{
    private static Scanner sc;
    private static PrintStream ps;
    
    /**
     * Do it!
     */
    private void doit()
    {
        char stones[] = sc.next().toCharArray();
        
        // The trick is very simple: It's possible iff there's the same number of Black stones as White stones.
        //
        // Suppose B=W. Then pick one White stone. All the others form a group with one fewer
        // white stone than black stones, and thus can be reduced to a single black stone. And, you're done.
        //
        // Now suppose B!=W. Each operation always removes the same number of W stones as B stones.
        // So, you can never erase the difference.
        int w=0, b=0;
        for( char stone : stones )
        {
            if( stone=='B' ) ++b;
            else if( stone=='W' ) ++w;
            else System.err.println( "PANIC!! Sonething other than B or W in input: " + stone );
        }
        
        ps.println( b==w ? 1 : 0 );
    }
        
    /**
     * main
     * 
     * @param args
     * @throws Exception
     */
    public static void main( String[] args ) throws Exception
    {
        sc = new Scanner( System.in );
        ps = System.out;
        
        new yinyang_vanb().doit();
    }

}