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 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace::std;
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
#define endl "\n"
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/numeric/interval.hpp>
#include <boost/numeric/interval/io.hpp>
#include <boost/any.hpp>
using namespace::std;
namespace mp = boost::multiprecision;
typedef mp::number<mp::cpp_dec_float<32,int64_t>> cfloat;
typedef mp::cpp_int cint;
typedef long long lint;
typedef long long ll;
typedef long double ldouble;
typedef vector<lint> vec;
typedef vector<vector<lint>> mat;
typedef vector<vector<vector<lint>>> mat3;
typedef vector<double> dvec;
typedef vector<vector<double>> dmat;
typedef vector<vector<vector<double>>> dmat3;
typedef vector<string> svec;
typedef vector<vector<string>> smat;
typedef vector<vector<vector<string>>> smat3;
typedef vector<pair<lint,lint>> pvec;
typedef vector<vector<pair<lint,lint>>> pmat;
typedef vector<vector<vector<pair<lint,lint>>>> pmat3;
#define rep(i, n) for(lint i = 0; i < (lint)(n); i++)
#define irep(i) for(lint i = 0;; i++)
#define irep1(i) for(lint i = 1;; i++)
#define irep2(i) for(lint i = 2;; i++)
#define rrep(i, n) for(lint i = (lint)(n-1); i >-1; i--)
#define rrepi(i,a,b) for(lint i = (lint)(b-1); i >a-1; i--)
#define repi(i,a,b) for(lint i=lint(a);i<lint(b);i++)
#define rep2(i,a,b,c) for(lint i=lint(a);i>lint(b);i+=c)
#define all(x) (x).begin(),(x).end()
#define PI 3.141592653589793
#define dist(x1,y1,x2,y2) (pow(pow(x2-x1,2)+pow(y2-y1,2),0.5))
#define output(v) do{bool f=0;for(auto i:v){cout<<(f?" ":"");if(i>INF/2)cout<<"INF";else cout<<i;f=1;}cout<<"\n";}while(0)
#define output2(v) for(auto j:v)output(j);
#define input(a,n) lint n;cin>>n;vector<lint>a(n);rep(i,n)cin>>a[i];
#define SUM(v) accumulate(all(v),0LL)
#define INF (1LL<<60)
#define IINF (1<<30)
#define LINF 9223372036854775807
#define MOD 1000000007
inline lint gcd(lint a,lint b){return b?gcd(b,a%b):a;}
inline lint lcm(lint a,lint b){return a*b/gcd(a,b);}
inline bool chmin(auto& s,const auto& t){bool res=s>t;s=min(s,t);return res;}
inline bool chmax(auto& s,const auto& t){bool res=s<t;s=max(s,t);return res;}
inline lint mex(vector<lint> s){
    lint res=0;
    sort(all(s));
    while(binary_search(all(s),res))res++;
    return res;
}
vector<lint> dx={-1,1,0,0,1,1,-1,-1};
vector<lint> dy={0,0,-1,1,1,-1,1,-1};
int main(){
    lint n;
    cin>>n;
    vec a(n);
    rep(i,n)cin>>a[i];
    rep(i,n-1)a[i+1]+=a[i];
    lint ans=INF;
    rep(i,n-1){
        lint mn=INF,mn2=INF;
        auto itr=lower_bound(a.begin(),a.begin()+i+1,a[i]*0.5);
        auto itr2=lower_bound(a.begin()+i+1,a.end(),(a[i]+a[n-1])*0.5);
        if(itr<a.begin()+i+1)chmin(mn,abs((*itr)*2-a[i]));
        if(itr>a.begin())chmin(mn,abs(a[i]-*(itr-1)*2));
        if(itr2<a.end())chmin(mn2,abs((*itr2)*2-(a[n-1]+a[i])));
        if(itr2>a.begin()+i+1)chmin(mn2,abs((a[n-1]+a[i])-*(itr2-1)*2));
        chmin(ans,max({(abs(a[i]*2-a[n-1])+mn+mn2)/2,mn,mn2}));
    }
    cout<<ans;
}