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 C++14 (Clang 3.8.0) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <list>
#include <limits>
#include <set>
#include <map>
#include <functional>

#include <inttypes.h>
#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

using namespace std;
typedef long long ll;

#ifndef PRId64
#define PRId64 "lld"
#endif
#ifndef SCNd64
#define SCNd64 "lld"
#endif
#define SC sc()
#define PT pr()
#define FORi(i,b,e) for (int i = (b), _ = (e); i < _; ++i)
#define FORe(i,b,e) for (int i = (b), _ = (e); i <= _; ++i)
#define FORre(i,b,e) for (int i = (b), _ = (e); i >= _; --i)

static int sc_ret = 0;
struct sc
{
    sc& operator()(char& v) { v = sc_ret = getchar(); return *this; }
    sc& operator()(int& v) { sc_ret = scanf("%d", &v); return *this; }
    sc& operator()(unsigned& v) { sc_ret = scanf("%u", &v); return *this; }
    sc& operator()(double& v) { sc_ret = scanf("%lf", &v); return *this; }
    sc& operator()(char* v) { sc_ret = scanf("%s", v); return *this; }
    sc& operator()(string& v) { sc_ret = (bool)(cin >> v); return *this; }
    sc& operator()(ll& v) { sc_ret = scanf("%" SCNd64, &v); return *this; }
    sc& ch(char& v) { sc_ret = scanf(" %c", &v); return *this; }
    sc& gets(char* v) { sc_ret = fgets(v, INT_MAX, stdin) != 0; v[strlen(v) - 1] = 0; return *this; }
    operator bool() const { return sc_ret > 0; }
    void read(int& v)
    {
        int x = 0, k = 1;
        char c = getchar();
        while (c < '0' || c > '9')
        {
            if (c == '-') k = -1;
            c = getchar();
        }
        while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c - 48), c = getchar();
        v = x * k;
    }
};
struct pr
{
    pr& ln() { putchar('\n'); return *this; }
    pr& operator()(char v) { putchar(v); return *this; }
    pr& operator()(int v) { printf("%d", v); return *this; }
    pr& operator()(double v) { printf("%.2f", v); return *this; }
    pr& operator()(const char* fmt, double v) { printf(fmt, v); return *this; }
    pr& operator()(const char* v) { printf("%s", v);return *this; }
    pr& operator()(string v) { printf("%s", v.c_str());return *this; }
    pr& operator()(ll v) { printf("%" PRId64, v);return *this; }

    void write(int v)
    {
        int cnt = 0; char c[15];
        if (v == 0)
        {
            putchar('0');
            return;
        }
        if (v < 0) putchar('-'), v = -v;
        while (v) c[++cnt] = (v % 10) + 48, v /= 10;
        while (cnt > 0) putchar(c[cnt--]);
    }
    template <typename T>
    void ln(T* arr, int size)
    {
        if (size > 0)
        {
            (*this)(arr[0]);
            for (int i = 1; i < size; ++i)
            {
                putchar(' ');
                (*this)(arr[i]);
            }
            putchar('\n');
        }
    }

    template <typename T>
    void muln(T* arr, int size)
    {
        for (int i = 0; i < size; ++i)
        {
            (*this)(arr[i]);
            putchar('\n');
        }
    }
};

const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const int maxn = 200000 + 10;
const int maxm = 100000 + 10;

ll a[maxn];

int main()
{
    int n, t, k;
    SC(n);
    FORe(i, 1, n)
    {
        int v;
        SC.read(v);
        a[i] = v + a[i - 1];
    }
    int l = 1, r = 3; ll ret = a[n];
    int ml = 2, mr = n - 1;
    //int ml = min(n - 2, max(2, (int)(lower_bound(a + 1, a + n, a[n] / 2) - a) - 100));
    //int mr = min(n - 1, ml + 200);
    l = max(1, (int)(lower_bound(a + 1, a + ml - 1, a[ml] / 2) - a) - 1);
    r = max(ml + 1, (int)(lower_bound(a + ml + 1, a + n, (a[n] - a[ml]) / 2 + a[ml]) - a) - 1);
    //l = 1; r = ml + 1;
    FORi(i, ml, mr)
    {
        for (; l < i - 1; ++l)
        {
            if (min(a[l], a[i] - a[l]) >= min(a[l + 1], a[i] - a[l + 1]))
                break;
        }
        for (; r < n - 1; ++r)
        {
            if (min(a[r] - a[i], a[n] - a[r]) >= min(a[r + 1] - a[i], a[n] - a[r + 1]))
                break;
        }
        ret = min(ret, max(max(a[r] - a[i], a[n] - a[r]), max(a[l], a[i] - a[l]))
            - min(min(a[l], a[i] - a[l]), min(a[r] - a[i], a[n] - a[r])));
    }
    PT(ret);
    return 0;
}