(define (powmod b p m)
  (cond
   ((zero? p) 1)
   ((zero? b) 0)
   (else
    (if (zero? (mod m 2))
        (letrec ((prev (powmod b (div p 2) m)))
          (mod (* prev prev) m))
        (letrec ((prev (powmod b (- p 1) m)))
          (mod (* b prev) m))))))

(define (answer m)
  (mod (+ (mod 2017 m)
          (powmod 2017 4034 m))
       m))

(letrec ((m (read)))
  (print (answer m)))