Problem:
Score : 100 points

Problem Statement

Taro and Jiro will play the following game against each other.

Initially, they are given a sequence a = (a_1, a_2, \ldots, a_N).
Until a becomes empty, the two players perform the following operation alternately, starting from Taro:

Remove the element at the beginning or the end of a. The player earns x points, where x is the removed element.

Let X and Y be Taro's and Jiro's total score at the end of the game, respectively.
Taro tries to maximize X - Y, while Jiro tries to minimize X - Y.

Assuming that the two players play optimally, find the resulting value of X - Y.

Constraints

All values in input are integers.

1 \leq N \leq 3000

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

Print the resulting value of X - Y, assuming that the two players play optimally.

Sample Input 1

4
10 80 90 30

Sample Output 1

10

The game proceeds as follows when the two players play optimally (the element being removed is written bold):

Taro: (10, 80, 90, 30) → (10, 80, 90)

Jiro: (10, 80, 90) → (10, 80)

Taro: (10, 80) → (10)

Jiro: (10) → ()

Here, X = 30 + 80 = 110 and Y = 90 + 10 = 100.

Sample Input 2

3
10 100 10

Sample Output 2

-80

The game proceeds, for example, as follows when the two players play optimally:

Taro: (10, 100, 10) → (100, 10)

Jiro: (100, 10) → (10)

Taro: (10) → ()

Here, X = 10 + 10 = 20 and Y = 100.

Sample Input 3

1
10

Sample Output 3

10

Sample Input 4

10
1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000 1

Sample Output 4

4999999995

The answer may not fit into a 32-bit integer type.

Sample Input 5

6
4 2 9 7 1 5

Sample Output 5

2

The game proceeds, for example, as follows when the two players play optimally:

Taro: (4, 2, 9, 7, 1, 5) → (4, 2, 9, 7, 1)

Jiro: (4, 2, 9, 7, 1) → (2, 9, 7, 1)

Taro: (2, 9, 7, 1) → (2, 9, 7)

Jiro: (2, 9, 7) → (2, 9)

Taro: (2, 9) → (2)

Jiro: (2) → ()

Here, X = 5 + 1 + 9 = 15 and Y = 4 + 7 + 2 = 13.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//
//  ROIGold.cpp
//  Main calisma
//
//  Created by Rakhman on 05/02/2019.
//  Copyright © 2019 Rakhman. All rights reserved.
//

#include <cstring>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <queue>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cassert>
#include <iterator>

#define ios ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
#define S second
#define F first
#define pb push_back
#define nl '\n'
#define NL cout << '\n';
#define EX exit(0)
#define all(s) s.begin(), s.end()
#define no_answer return cout << "NO", 0;
#define FOR(i, start, finish, k) for(int i = start; i <= finish; i += k)

const int MXN = 4e5 + 200;
const long long MNN = 4e2 + 200;
const long long MOD = 1e9 + 7;
const long long INF = 1e13;
const int OO = 1e9 + 500;

typedef long long llong;
typedef unsigned long long ullong;

using namespace std;

int n, k, a[MXN];
llong dp[MNN][MNN][2];

llong get(llong action, llong l, llong r){
    if(dp[l][r][action] != -INF) return dp[l][r][action];
    if(l == r) {
        if(action == 1) return a[l];
        else return -a[l];
    }
    if(action == 1){
        llong choice1 = get(0, l + 1, r) + a[l];
        llong choice2 = get(0, l, r - 1) + a[r];
        return dp[l][r][1] = max(choice1, choice2);
    }else{
        llong choice1 = get(1, l + 1, r) - a[l];
        llong choice2 = get(1, l, r - 1) - a[r];
        return dp[l][r][0] = min(choice1, choice2);
    }
}

int main () {
    ios;
    cin >> n;
    for(int i = 1; i <= n; i++){
        for(int j = i; j <= n; j++){
            dp[i][j][0] = dp[i][j][1] = -INF;
        }
    }
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    cout << get(1, 1, n);
}