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: Accepted)

/*
    STARK_BOY
*/
#include <bits/stdc++.h>
#include <fstream>
using namespace std;
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repe(i,a,b) for(ll i=a;i<=b;i++)
#define bac(i,a,b) for(ll i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long ll;
typedef vector< ll > vi;
typedef vector<pair<ll,ll> > vpii;
typedef pair<ll,ll> pii;
#define all(c) c.begin(),c.end()
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define setbitc(x) __builtin_popcount(x)
#define init(x,a) memset(x,a,sizeof(x))
#define PI           3.14159265358979323846  /* pi */
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
#define N 3005
//Variables Start
ll inf=1e18,MOD=1e9+7;
ll n,m,k,ar[N],dp[N][N][2];
string s,s1;
//Variables End

ll compute(ll l,ll r,ll st)
{
	if(dp[l][r][st]!=-inf)return dp[l][r][st];
	if(l==r && st)return ar[l];
	if(l==r && !st)return -ar[l];

	if(st)
	{
		dp[l][r][st]=max(compute(l+1,r,1-st)+ar[l],compute(l,r-1,1-st)+ar[r]);
	}
	else
	{
		dp[l][r][st]=min(compute(l+1,r,1-st)-ar[l],compute(l,r-1,1-st)-ar[r]);
	}

	return dp[l][r][st];
}
int main()
{
    ios
    int ti=1;
    //cin>>ti;
    while(ti--)
    {
		cin>>n;
		rep(i,0,n+1)rep(j,0,n+1)rep(k,0,2)dp[i][j][k]=-inf;

		rep(i,0,n)cin>>ar[i];
		cout<<compute(0,n-1,1);
    }
    return 0;
        
}