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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
    n,i:longint;
    a:array[0..200010] of longint;
    b:array[0..200010] of int64;
    min,max,ans:int64;
procedure find(h,t:longint);
var
    l,r,q,mid,x,y,x1,y1,x2,y2:longint;
begin
    l:=h;r:=t;q:=(b[t]-b[h-1]) div 2;
    while l<>r-1 do
        begin 
            mid:=(l+r) div 2;
            if b[mid]-b[h-1]>q then r:=mid else l:=mid;
        end;
    x1:=b[l]-b[h-1];
    y1:=b[t]-b[l];
    x2:=b[r]-b[h-1];
    y2:=b[t]-b[r];
    if abs(x1-y1)<abs(x2-y2) then 
        begin
            x:=x1;y:=y1;
        end
    else
        begin
            x:=x2;y:=y2;
        end;
    if x>max then max:=x;
    if x<min then min:=x;
    if y>max then max:=y;
    if y<min then min:=y;
end;
begin
    readln(n);
    for i:=1 to n do
        begin
            read(a[i]);
            b[i]:=b[i-1]+a[i];
        end;
    ans:=maxlongint;
    for i:=2 to n-2 do
        begin
            max:=0;
            min:=maxlongint;
            find(1,i);
            find(i+1,n);
            if max-min<ans then ans:=max-min;
        end;
    writeln(ans);
end.