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)

//****** @mdazmat9 **********
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define UB upper_bound
#define LB lower_bound
#define BS binary_search
#define EB emplace_back
#define PB push_back
#define endl "\n"
#define MOD2 1000000007
#define MOD 998244353
#define F first
#define S second
#define ALL(a) (a).begin(),(a).end()
typedef pair<int, int> pr;
typedef vector<int> VI;
typedef vector<pr> VP;
typedef vector<string> VS;
typedef vector<vector<int>> VV;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define trace1(x)                cout<<#x<<": "<<x<<endl
#define trace2(x, y)             cout<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cout<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cout<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
#define trace(v) for(auto it=v.begin();it!=v.end();it++)cout<<*it<<" ";cout<<endl;
int fast_pow(int x, int y, int p);
const int MAX_SIZE = 200001;
int n;VI a;
int dp[3001][3001];
int rec(int i,int j){
    if(i>j)return 0;
    int& ans=dp[i][j];
    if(ans!=-1)return ans;
    int op1=a[i]-rec(i+1,j);
    int op2=a[j]-rec(i,j-1);
    ans=max(op1,op2);
    return ans;
}
void solve() {
    cin>>n;
    for(int i=0;i<n;i++){
        int x;cin>>x;
        a.EB(x);
    }
    memset(dp,-1, sizeof(dp));
    cout<<rec(0,n-1);
}

int32_t main() {
    IOS;
    int test = 1;
//    cin >> test;
    for(int i=1;i<=test;i++){
//        cout<<"Case #"<<i<<": ";
        solve();
    }
    return 0;
}
int fast_pow(int x, int y, int p) {
    int res = 1;
    x = x % p;
    while (y > 0) {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}