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++ (GCC 9.2.1) (judge verdict: Accepted)

/* ▄▄▄▄▄▄▄▄▄▄▄  ▄▄▄▄▄▄▄▄▄▄▄▄            ▄            ▄▄▄▄▄▄▄▄▄▄▄  ▄         ▄       ▄▄▄▄▄▄▄▄▄▄▄  ▄▄▄▄▄▄▄▄▄▄▄  ▄▄▄▄▄▄▄▄▄▄
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//#pragma optimization_level 3

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <ctime>
#include <queue>
#include <math.h>
#include <deque>
#include <stack>
#include <iomanip>
#include <assert.h>
#include <stdio.h>
#include <cstring>
#include <random>

using namespace std;

#define int long long
#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair<int, int>
#define pld pair<ld, ld>
#define vi vector <int>
#define vpi vector <pii>
#define vld vector <ld>
#define try tr
#define left left11
#define right right11
#define lb lower_bound
#define ub upper_bound
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define pf push_front
#define sqr(a) ((a) * (a))
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define x first
#define y second

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

clock_t START;

inline ld TIME() {
    return (ld) (clock() - START) / CLOCKS_PER_SEC;
}

ld TL = 2.0; 
int calls = 0;

inline bool IS() {
    if(++calls == 1000) {
        calls = 0;
        if(TL - TIME() < 0.1)
            return true;
    }
    return false;
}

template<typename T1, typename T2>
inline void amin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void amax(T1 &a, T2 b) { if (a < b) a = b; }

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << p.first << ' ' << p.second << endl;
    return os;
}

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (auto &u : v) is >> u;
    return is;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
    for (auto &u : v) os << u << ' ';
    os << endl;
    return os;
}

int hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937}, mod = hash_mod[rnd() % 4];
int hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];

int binpow(int x, int p) {
    int res = 1;
    while(p) {
        if(p % 2)
            res = (res * x) % mod;
        x = (x * x) % mod;
        p /= 2;
    }
    return res;
}

const int N = 1e6 + 7, NS = 3e3 + 7, lg = 20, sq = 550, inf = 1e18 + 7, SM = 1e2 + 7;
const ld eps = 1e-9;
int a[N], n, d[NS][NS];
int turn(int l, int r) {
	return (n - (r - l + 1)) % 2;
}
int solve(int l, int r) {
	if(d[l][r])
		return d[l][r];
	if(l == r)  {
		if(!turn(l, r))
			d[l][r] = a[l];
		else
			d[l][r] = -a[l];
		return d[l][r];
	}
	if(!turn(l, r)) 
		d[l][r] = max(a[l] + solve(l + 1, r), a[r] + solve(l, r - 1));
	else
		d[l][r] = min(solve(l + 1, r) - a[l], solve(l, r - 1) - a[r]);
	return d[l][r];
}
void solve() {
	cin >> n;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		sum += a[i];
	}
	cout << solve(0, n - 1);
}

signed main() {
    fast
    solve();
}