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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

class Main{
	static int G;
	static int D;
	static int[] p;
	static int[] c;
	static boolean[] used;
	static long ans = 1000000000;


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int n = sc.nextInt();
		long[] a = new long[n+1];
		for(int i = 1;i <= n;i++){
			a[i] = sc.nextLong();
		}
		long[] sum = new long[n+1]; 
		for(int i = 1;i <= n;i++){
			sum[i] = a[i] + sum[i-1];
//			System.out.print(sum[i] + " ");
		}
//		System.out.println();
		long ans = 1000000000;
		for(int i = 2;i <= n-2;i++){				//真ん中の仕切りのいち
//			System.out.println("ans "+ans);
			long left = lowerbond(sum[i]/2, sum,0,1,i);			//左の仕切りのいち
			//System.out.println("left "+left);

			long A = sum[(int)left];
			long B = sum[i] - sum[(int)left];
			
			long right = lowerbond((sum[n]-sum[i])/2, sum,sum[i],i+1,sum.length);			//右の仕切りのいち
			long C = sum[(int)right] - sum[i];
			long D = sum[n] - sum[(int)right];

//			System.out.println("left "+left+" right "+right);
			Long[] hako = {A,B,C,D};
			Arrays.sort(hako);
			ans = Math.min(ans, hako[3]-hako[0]);
//			System.out.println();

		}
		
		System.out.println(ans);
		


	}
	static long lowerbond(long k, long[] data,long d,int min,int max){		//kより小さい最小の値をdataの中から探し、返す二分探索
//		int min = -1;
//		int max = data.length;

		while(max-min > 1){
			int mid = (max+min)/2;
//			System.out.println("max "+max + " mid "+mid+" min "+min);

			if(data[mid]-d >= k){
				max = mid;
			}else{
				min = mid;
			}
		}
//		System.out.println("ok");

		return min;
	}



}

class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return from - otherpair.from;
	}
}