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)

//  Mayank Agarwal
//  IIIT Hyderabad



#include<bits/stdc++.h>
using namespace std;
const long long int M = 1000000007; 
#define ll long long int
#define ld double
#define pb push_back
#define mp make_pair
ll power(ll x, ll y,  ll p) 
{ 
    ll res = 1;      
    x = x % p;   
    while (y > 0) 
    { 
        if (y & 1) 
            res = (res*x) % p; 
        y = y>>1;x = (x*x) % p; 
    } return res%p; 
} 
ll modinver(ll a,ll b)
{
    return (power(a,1,M)*power(b,M-2,M))%M;
}
void SieveOfEratosthenes(int n) 
{   
    bool prime[10005]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
} 
ll dp[3003][3003];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    ll n;
    cin>>n;
    std::vector<ll> v(n);
    for(ll&x:v)
    {
        cin>>x;
    }
    for (int i = 0; i < n; ++i)
    {
        dp[i][i]=v[i];
    }
    for (int i = n-1; i>=0; --i)
    {
        for (int j = i+1; j<n ; ++j)
        {
            dp[i][j]=max(v[i]-dp[i+1][j],-dp[i][j-1]+v[j]);
        }
    }
    cout<<dp[0][n-1]<<'\n';
}