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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
    #define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5
#define PI 3.141592653589793238

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
void cin(ll *n){ scanf("%lld",&(*n)); }
void cin2(ll *a, ll *b){ scanf("%lld%lld",&(*a),&(*b)); }
void cin3(ll *a, ll *b, ll *c){ scanf("%lld%lld%lld",&(*a),&(*b),&(*c)); }
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
double POW_d(double a, double b){double c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod_MOD1(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
ll mod_p(ll n ,ll p){n+= n<0?((-n)/p+1)*p:0; return n%=p;}
ll change_into_num(char s[] , ll len, ll p){ return !p ? 0 : POW(10,p-1)*(s[len-p]-'0') + change_into_num(s,len,p-1); }

void lr_lower( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am < val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am <= val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
void lr_upper( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am <= val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am < val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
int cmp_lower( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a>=b ? 1 : 0 ) : ( a>b ? 1 : 0 ) ; }
int cmp_upper( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a<=b ? 1 : 0 ) : ( a<b ? 1 : 0 ) ; }
// return smallest p  which meets  a[p]==val :1  >=:2   >:3
ll lower_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_lower(&l,&r,a[ (l+r)/2 ],val,type);  return cmp_lower(a[l],val,type) ? l : cmp_lower(a[r],val,type) ? r : -1;  }
// return biggest p  which meets   a[p]==val :1  <=:2   <:3
ll upper_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_upper(&l,&r,a[ (l+r)/2 ],val,type); return cmp_upper(a[r],val,type) ? r : cmp_upper(a[l],val,type) ? l : -1; }
// count i  which meets ai==x
ll count(ll a[], int l, int r, ll x){  int p = lower_bound(a,l,r,x,1);  return p==-1 ? 0 : upper_bound(a,p,r,x,1)-p+1; }

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
int cmp_string( const void * a , const void * b ) {  return strcmp( (char *)a , (char *)b ); }  // qsort((void*)s,n,sizeof(s[0]),int_sort );
int cmp_char(const void * a, const void * b) { return *(char *)a - *(char *)b;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
void sort_string(int n,int size,char s[][size]){ qsort( (void*)s , n , sizeof(s[0]) , cmp_string ); }
void sort_char(char *s){ qsort( (void *)s , strlen(s) , sizeof(char) , cmp_char ); }
ll unique_string(ll n ,ll size, char s[][size]){ ll ans=1; rep(i,1,n) if( strcmp(s[i],s[i-1]) ) ans++; return ans; }
ll unique_num(ll n , ll a[]){ ll ans=1; rep(i,1,n) if( a[i]!=a[i-1] ) ans++; return ans; }

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}



int main(void){
    // fgets(s,sizeof(s),stdin); 
    // char ;
    ll n;
    ll ans=INF;
    cin(&n);
    // scanf("%s",);
    ll a[n];
    rep(i,0,n){
        cin(&a[i]);
    }

    ll sum[n+1];
    sum[0]=0;
    rep(i,0,n) sum[i+1]=sum[i]+a[i];
    // printf("%lld\n",sum[n]);
    // rep(i,0,n+1) printf("%lld ",sum[i]);
    // PN;

    rep(i,1,n-2){
        ll l1=2,r1=i+1,l2=i+2,r2=n-1;
        ll piv1[2],piv2[2];
        piv1[0]=upper_bound(sum,l1,r1,sum[i+1]/2,2);
        piv1[1]=lower_bound(sum,l1,r1,sum[i+1]/2,2);
        piv2[0]=upper_bound(sum,l2,r2,(sum[n]-sum[i+1])/2+sum[i+1],2);
        piv2[1]=lower_bound(sum,l2,r2,(sum[n]-sum[i+1])/2+sum[i+1],2);
        rep(j,0,2){
            if(piv1[j]==-1) continue;
            rep(k,0,2){
                if(piv2[k]==-1) continue;
                ll num[4];
                num[0]=sum[piv1[j]-1];
                num[1]=sum[i+1] - sum[piv1[j]-1];
                num[2]=sum[piv2[k]] - sum[i+1];
                num[3]=sum[n] - sum[piv2[k]];
                sortup(num,4);
                ans = min2(ans, ABS(num[0]-num[3]) );
            }
        }
        // printf("%lld\n",i);
        // printf("%lld %lld %lld %lld\n",piv1[0],piv1[1],piv2[0],piv2[1]);
        // printf("||%lld||\n",ans );
    }

    
    printf("%lld\n",ans);
    return 0;
}