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>
#include <ext/pb_ds/assoc_container.hpp>
/*#include <boost/multiprecision/cpp_int.hpp> 
using boost::multiprecision::cpp_int;*/
using namespace __gnu_pbds;
using namespace std;

#define ff              first
#define ss              second
#define int              long long int
#define pb              push_back
#define mp12            make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pq1             priority_queue<int>
#define pqr1            priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zerobits(x)      __builtin_ctzll(x)
#define mod             1000000007
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
// mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;


void fast()
{
    //ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

}
void file(){
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
}

int modInv(int a,int b,int m)
{
    int res=1;
    a=a%m;
    while(b>0){
        if(b&1){
            res=(res*a)%m;
        }
        a=(a*a)%m;
        b=b>>1;
    }
    return res;
}


int p[1000006];
vi prime;
//int mrp[1000000];
void seive(){
    for(int i=2;i<1000006;i++) {
        if(p[i]){
            prime.pb(i);
            for(int j=i*i;j<=1000006;j=j+i){
                p[j]=0;
            }
        }
    }

}
// int c;
int dp[3001][3001];
int solve(int i,int j,int a[]){
    if(i>j) return 0;
    if(dp[i][j]!=-1) return dp[i][j];

    int op1=0;
    int op2=0;
    op1=a[i]+min((solve(i+2,j,a)),(solve(i+1,j-1,a)));
    op2=a[j]+min((solve(i+1,j-1,a)),(solve(i,j-2,a)));
    return dp[i][j]=max(op1,op2);
}


int32_t main()
{
    // memset(prime,0,sizeof(prime));
    //memset(p,1,sizeof(p));   
    //seive();
   // file();
    fast();
    //int mo=1000000007;
   int n;
   cin>>n;
   int a[n];
   int s=0;
   for(int i=0;i<n;i++) {
        cin>>a[i];
        s+=a[i];
   }
    memset(dp,-1,sizeof(dp));
    int X=solve(0,n-1,a);
    int Y=s-X;
    cout<<(X-Y);



    return 0;
    
    
}