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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define fi first
#define sec second 
#define MOD 1000000007
#define pb push_back
#define endl "\n"
#define test cout<<"debuged\n";
#define INF 10000000000
#define si 1000000
#define pi pair<ll,ll>

ll dp[3010][3010];
ll vis[3010][3010];
ll solve(ll s,ll e,vector<ll> v,ll n)
{
  if(s>e)
  {
    return 0;
  }
 if(vis[s][e]==1)
    return dp[s][e];
  ll d=n-(e-s+1);
  vis[s][e]=1;
  if(d%2==0)
  {
    ll h=(-1*INF);
    ll k= max(solve(s+1,e,v,n)+v[s],solve(s,e-1,v,n)+v[e]);
     dp[s][e]=max(k,h);
    return dp[s][e];
  }
  else
  {
    ll h=(INF);
    ll a= solve(s+1,e,v,n)-v[s];
    ll b= solve(s,e-1,v,n)-v[e];
    a=min(a,h);
    dp[s][e]=min(a,b);
    return dp[s][e];
  }

}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    ll n,k,s=0;
    cin>>n;
    vector<ll> v(n,0);
    for(auto &x:v)
      {
         cin>>x;
         s+=x;
      }
   
    for(ll i=0;i<n+5;i++)
      for(ll j=0;j<n+5;j++)
        vis[i][j]=0;

    ll p=solve(0,n-1,v,n);  
    cout<<p;
}