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)

    #include<bits/stdc++.h>
    using namespace std;
     
     
    #define ll long long
    #define pb push_back
    #define mp make_pair
    #define loop(i, a, b) for(int i = (a); i < (b); i++)
    #define loopb(i, a, b) for(int i = (a); i > (b); --i)
     
    typedef pair <int,int> ii;
    typedef vector <int> vi;
    typedef vector <ii> vii;
    typedef vector <ll> vl;
     
    const int mod = 1e9+7;
    const ll inf = 2e18+5;
     
    int main()
    {
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);

        int n;
        cin>>n;
        vl a(n);
        loop(i,0,n) cin>>a[i];

        ll dp[3005][3005]; //dp[i][j] is the answer when interval of [i,j] is remaining. So our final answer will be [0,n-1]

        for(int i=n-1; i>=0; i--){
            for(int j=i; j<=n-1; j++){
                if(i==j)
                    dp[i][j] = a[i];
                else
                    dp[i][j] = max(a[i] - dp[i+1][j] , a[j] - dp[i][j-1]);    
            }
        }

        cout<<dp[0][n-1];

        return 0;

}