/* Problem 4--Great Circle Trajectory
   This was intended to be the hardest problem in the set, requiring a
   thorough knowledge of trigonometry.  Note that the ACM contest almost
   always have a trigonometric problem.

   If you are dismayed that there is a "magic formula" that solves the
   problem that you may not have known, well, I'm as surprised as you.
   I didn't know there was a magic formula until I worked one out while
   writing this solution program. */

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define PI 3.14159265358979323846

FILE *in, *out;
double slat, slong, dlat, dlong;
char slatdir, slongdir, dlatdir, dlongdir;

int main (int argc, char **argv);
void Process (void);

int main (int argc, char **argv) {

 in = fopen ("prob4.in","r");
 out = fopen ("prob4.out","w"); /* read in data */
 while (fscanf (in,"%lf",&slat),slat >= 0) {
  fscanf (in,"%1s %lf %1s %lf %1s %lf %1s",&slatdir,&slong,&slongdir,
             &dlat,&dlatdir,&dlong,&dlongdir);
  Process ();
 }
 fclose (in);
 fclose (out);
 return EXIT_SUCCESS;
}

/* Process processes a test case. */
void Process (void) {

  double traj;

 if (slatdir=='S') slat = -slat; /* Negate angle if necessary */
 if (dlatdir=='S') dlat = -dlat;
 if (slongdir=='W') slong = -slong;
 if (dlongdir=='W') dlong = -dlong;
 slat *= PI/180;
 slong *= PI/180; /* Convert to radians */
 dlat *= PI/180;
 dlong *= PI/180; /* Magic formula */
 traj = atan2 (cos(dlat)*sin(dlong-slong),
              cos(slat)*sin(dlat)-sin(slat)*cos(dlat)*cos(dlong-slong));
 traj *= 180/PI; /* convert to degrees */
 if (fabs (traj) < 0.005) traj = 0;
 if (traj < 0) fprintf (out,"%.2f degrees west of north\n",-traj);
 else fprintf (out,"%.2f degrees east of north\n",traj);
}