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)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define Ma7moud_7amdy                 \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define Open_Sesame Open()
#define all(v) ((v).begin()), ((v).end())
#define allr(v) ((v).rbegin()), ((v).rend())
#define clr(arr, x) memset(arr, x, sizeof arr)
#define endl "\n"
#define watch(x) cout << #x << " = " << x << endl;
#define RT(x) return cout << (x), 0;
#define Accepted 0
typedef long long ll;
typedef vector<int> vi;
const int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
const int dy[] = { 0, 1, 0, -1, 1, -1, 1, -1 };
void Open()
{
#ifndef ONLINE_JUDGE 
	freopen("Input.txt", "r", stdin);
	freopen("Output.txt", "w", stdout);
#endif
}
const int mod = ll(1e9 + 7), N = 3000 + 5;
//“Nobody but you have to believe in your dreams to make them a reality.” ― Germany Kent
ll arr[N], n, mem[N][N][2];
ll solve(int i, int j, bool state) {
	if (i > j)return 0;
	ll& ret = mem[i][j][state];
	if (~ret)return ret;
	if (state) {
		ret = solve(i + 1, j, !state) + arr[i];;
		ret = max(ret, solve(i, j - 1, !state) + arr[j]);
	}
	else {
		ret = solve(i + 1, j, !state) + -arr[i];;
		ret = min(ret, solve(i, j - 1, !state) + -arr[j]);
	}

	return ret;
}
int main()
{
	Ma7moud_7amdy;
	Open_Sesame;
	clr(mem, -1);
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> arr[i];
	}
	cout << solve(0, n - 1, 1);
}