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)

#include <bits/stdc++.h>
#define lli long long int
#define vec2 vector<vector<lli>> 
#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define FRE freopen("explicit.in","r",stdin);freopen("explicit.out","w",stdout)
#define mp make_pair
#define pll pair <lli,lli>
#define vll vector <lli>
#define pq2 priority_queue<pair<lli,lli>>
#define f first
#define s second
#define pb push_back
#define pf push_front
#define minii LLONG_MAX
#define ld long double 
const long double pi = 3.1415926535898;
#define dec greater<lli>()
//const lli mod=1000000007;
const lli mod=100000000;
using namespace std;

int main()
{
    fio;
    lli n;
    cin>>n;
    lli a[n];
    for (int i=1;i<=n;i++)
    cin>>a[i];
    lli dp[3005][3005];
    memset(dp,0,sizeof(dp));
    for (int i=n;i>=1;i--)
    {
        for (int j=i;j<=n;j++)
        {
            dp[i][j]=max(a[i]-dp[i+1][j],a[j]-dp[i][j-1]);
        }
    }
    cout<<dp[1][n];
    return 0;
}