#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
// int mod = 998244353;
int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { a += b % mod; if (a >= mod) a -= mod; }
template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }

void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(12); // debuged
    int T;
    T = 1;
    while (T--) solve();
}

// https://atcoder.jp/contests/arc100/submissions/15483685
// 15
// 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1000
// のような例で, 真ん中を総和の半分らへんにするとほぼ右端になる
// 1 1 1 1 1    1 1 1 1 1    1 1 1 1    1000
// が最適
void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    cin >> a;
    
    vector<ll> acc(n + 1);
    rep(i, n) acc[i + 1] = acc[i] + a[i];
    
    ll ans = INF;
    int b = 2;
    rep3(bi_, 2, n - 1) { // 全探索!!
        if (bi_ <= 1 || bi_ >= n - 1) continue;
        
        // 前半を2分割
        int it2 = lb(acc, acc[bi_] / 2);
        rep3(qu_, it2 - b, it2 + b + 1) { // 前半後半の始め位置
            if (qu_ <= 0 || qu_ >= bi_) continue;
            
            int it22 = lb(acc, acc[bi_] + (acc[n] - acc[bi_]) / 2);
            rep3(qu_2, it22 - b, it22 + b + 1) { // 後半後半の始め位置
                if (qu_2 <= bi_ || qu_2 >= n) continue;
                
                vector<ll> v;
                v.eb(acc[qu_]);
                //(0+...+(bi_-1)) - (0+...+(qu_-1)) = (qu_+...+(bi_-1))
                v.eb(acc[bi_] - acc[qu_]);
                v.eb(acc[qu_2] - acc[bi_]);
                v.eb(acc[n] - acc[qu_2]);
                sort(all(v));
                chmin(ans, v[3] - v[0]);
            }
        }
    }
    cout << ans << endl;
}



Measured memory_bytes: 5488