Problem:
Score : 600 points

Problem Statement

Snuke has an integer sequence A of length N.

He will make three cuts in A and divide it into four (non-empty) contiguous subsequences B, C, D and E.
The positions of the cuts can be freely chosen.

Let P,Q,R,S be the sums of the elements in B,C,D,E, respectively.
Snuke is happier when the absolute difference of the maximum and the minimum among P,Q,R,S is smaller.
Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

Constraints

4 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

Sample Input 1

5
3 2 4 1 2

Sample Output 1

2

If we divide A as B,C,D,E=(3),(2),(4),(1,2), then P=3,Q=2,R=4,S=1+2=3.
Here, the maximum and the minimum among P,Q,R,S are 4 and 2, with the absolute difference of 2.
We cannot make the absolute difference of the maximum and the minimum less than 2, so the answer is 2.

Sample Input 2

10
10 71 84 33 6 47 23 25 52 64

Sample Output 2

36

Sample Input 3

7
1 2 3 1000000000 4 5 6

Sample Output 3

999999994

Solution in Common Lisp (SBCL 1.1.14) (judge verdict: Accepted)

(defun get-max (data start end)
  (if (zerop start)
      (aref data (1- end))
      (- (aref data (1- end))
         (aref data (1- start)))))

(defun get-diff (data start-point second-point end-point)
  (abs (- (get-max data second-point end-point)
          (get-max data start-point second-point))))

(defun equal-cut ()
  (declare (optimize (speed 3) (safety 0) (debug 0) (space 0)))
  (let* ((n (read))
         (data (make-array n :initial-contents (loop repeat n
                                                     sum (read) into summed-a
                                                     collect summed-a))))
    (loop for second-separator from 2 below (1- n)
          with first-separator = 1
          with third-separator = 3
          with minimized-diff = #.(* (* 2 (expt 10 5)) (expt 10 9))
          when (>= second-separator third-separator)
            do (setq third-separator (1+ second-separator))
          do (loop (if (and (< first-separator (1- second-separator))
                            (< (get-diff data 0 (1+ first-separator) second-separator)
                               (get-diff data 0 first-separator second-separator)))
                       (incf first-separator)
                       (return)))
             (loop (if (and (< third-separator (1- n))
                            (< (get-diff data second-separator (1+ third-separator) n)
                               (get-diff data second-separator third-separator n)))
                       (incf third-separator)
                       (return)))
             (let* ((max-list (list (get-max data 0 first-separator)
                                    (get-max data first-separator second-separator)
                                    (get-max data second-separator third-separator)
                                    (get-max data third-separator n)))
                    (diff (- (apply #'max max-list) (apply #'min max-list))))
               (if (< diff minimized-diff)
                   (setq minimized-diff diff)))
          finally (princ minimized-diff))))

(equal-cut)