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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <iomanip>
#include <string>
#include <deque>
#include <stack>
#include <cmath>

using namespace std;

#define PB push_back
#define MP make_pair
#define f first
#define s second
#define FOR(i,a,b) for (LL i = a; i < b; ++i)
#define RFOR(i, b, a) for (int i = b-1; i>=a; --i)
#define FILL(a, b) memset(a, b, sizeof a)
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<LL> VI;
typedef pair<LL, LL> PLL;

const int INF = 1000 * 1000 * 1000;
const LL LINF = 1LL * INF*INF;
const int MAX = 400040	;
const long double PI = acos(-1.);
const double EPS = 1e-6;
LL mod = 1000LL * 1000 * 1000 + 7;

int n,k;
LL arr[300];
LL dpT[3007][3007];
LL dp(int l, int r)
{
	if(l>r)
		return 0;
	if(l==r)
		return arr[l];
	if(dpT[l][r]!=-1LL)
		return dpT[l][r];
	LL ans = -LINF;
	ans=max(ans,arr[l]+min(dp(l+2,r)-arr[l+1],dp(l+1,r-1)-arr[r]));
	ans=max(ans,arr[r]+min(dp(l+1,r-1)-arr[l],dp(l,r-2)-arr[r-1]));
	return dpT[l][r] = ans;
}



int main()
{
	//freopen("input.tXXt", "r", stdin);
	ios_base::sync_with_stdio(0);
	FILL(dpT,-1);
	cin >> n;
	FOR(i,0,n)
		cin >> arr[i];
	cout<<dp(0,n-1);
	return 0;
}