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>

//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

#define ss second
#define ff first
#define int long long
//#define double long double
#define pb push_back


const int N = 2e5 + 51 + 51, MOD = 1e9 + 7, mod = 998244353, INF = 3e18;

mt19937 Flash_new_51(51515);

using vi = vector <int>;
using vp = vector <pair <int, int>>;

//gp_hash_table<int, int> table;

int a[N], dp[3000][3000];
bool ok[3000][3000];

signed main() {
    #ifdef Flash
     freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (n % 2 == 0) dp[i][i] = -a[i];
        else dp[i][i] = a[i];
    }
    for (int i = 1; i <= n; i++) {
        int len = n - i, kof = 1;
        if (len % 2 == 0) kof = -1;

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