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)

#define _CRT_SECURE_NO_WARNINGS
/*                   وتزودوا فإن خير الزاد التقوى                    */
/*       وما تدري نفس ماذا تکسب غدا وما تدري نفس باي ارض تموت      */
#include <bits/stdc++.h>
#include<unordered_set>
using namespace std;
#define endl '\n'
#define ll long long
#define PI acos(-1)
#define watch(x) cerr<<#x<<" = { "<<x<<" }\n"
#define sz(n) n.size()
#define Mo7med_Sayed fast();
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define RT(v) return cout<<v,0;
const double EPS = (1e-7);
const ll OO = INT_MAX, mod = 1e9 + 7;
void run() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
#else
#endif
}
void fast() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }
ll n, m, k, t;
vector<ll>v;
ll dp[3005][3005][2];
ll solve(int x,int y,bool b) {
	if (x>y)return 0;
	ll& ret = dp[x][y][b];
	if (~ret)return ret;
	if (b) {
		ret = max(solve(x + 1, y, b ^ 1) + v[x], solve(x , y-1, b ^ 1) + v[y]);
	}
	else {
		ret= min(solve(x + 1, y, b ^ 1) , solve(x , y-1, b ^ 1));
	}
	return ret;
}

int main()
{
	run();
	Mo7med_Sayed;

	cin >> n;
	v.resize(n);
	memset(dp, -1, sizeof dp);
	ll sum = 0;
	for (int i = 0; i < n; i++) {
		cin >> v[i] ; sum += v[i];
	}
	cout << solve(0, n - 1, 1)*2-sum;

}