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 Ruby (2.3.3) (judge verdict: Accepted)

n = gets.chomp.to_i
a = gets.chomp.split("\s").map(&:to_i)

sum = Array.new
a.each do |item|
  begin
    sum.push sum.last + item
  rescue
    sum.push item
  end
end


result = Array.new

# 左の最適値を計算
l = 0
c = 1
f_l = Array.new
diff = sum[-1]
while true do
    tmp_diff = (sum[c] - 2*sum[l]).abs

    #puts "l: #{l}, c: #{c}, diff: #{diff}, tmp_diff: #{tmp_diff}"

    if tmp_diff > diff then
        #puts "f_l[#{c}]: #{l-1}"
        f_l[c] = l-1
        l -= 1
        
        if c < n-2 then
            c += 1
            diff = sum[-1]
        else
            break
        end
    elsif l == c-1 then
        #puts "f_l[#{c}]: #{l}"
        f_l[c] = l
        
        if c < n-2 then
            c += 1
            diff = sum[-1]
        else
            break
        end
    elsif l < c-1 then
        diff = tmp_diff
        l += 1
    else
        break
    end
end

#p f_l

# 右の最適値を計算
r = 2
c = 1
f_r = Array.new
diff = sum[-1]
while true do
    tmp_diff = (sum[-1] - 2*sum[r] + sum[c]).abs

    #puts "c: #{c}, r: #{r}, diff: #{diff}, tmp_diff: #{tmp_diff}"

    if tmp_diff > diff then
        #puts "f_r[#{c}]: #{r-1}"
        f_r[c] = r-1
        r -= 1
        
        if c < n-2 then
            c += 1
            diff = sum[-1]
        else
            break
        end
    elsif r == n-1 then
        #puts "f_r[#{c}]: #{r}"
        f_r[c] = r
        
        if c < n-2 then
            c += 1
            diff = sum[-1]
        else
            break
        end
    elsif r < n-1 then
        diff = tmp_diff
        r += 1
    else
        break
    end
end

#p f_r

result = Array.new
for c in 1...n-2 do
    l = f_l[c]
    r = f_r[c]

    v1 = sum[l]
    v2 = sum[c] - sum[l]
    v3 = sum[r] - sum[c]
    v4 = sum[-1] - sum[r]

    tmp = [v1, v2, v3, v4]
    #puts "c: #{c}"
    #p tmp
    
    result.push tmp.max - tmp.min
end

puts result.min