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)

//
// Created by Hideaki Imamura on 2020-03-17.
//
# include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair <ll, ll> l_l;
typedef pair<int, int> i_i;

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

# define EPS (1e-7)
# define INF (1e9)
# define PI (acos(-1))
//const ll mod = 1000000007;

int N;
ll a[3100];
ll dp[3100][3100];

int main() {
    cin >> N;
    for (int i = 0; i < N; ++i) cin >> a[i];

    // dp[i][j] = 元の数列から連続する区間[i,j)をとった局面からスタートした時、X-Yの値
    for (int i = 0; i < N; ++i) dp[i][i] = 0;

    for (int len = 1; len <= N; ++len) {
        for (int i = 0; i + len <= N; ++i) {
            int j = i + len;
            if ((N - len) % 2 == 0) {
                dp[i][j] = dp[i+1][j] + a[i];
                chmax(dp[i][j], dp[i][j-1] + a[j-1]);
            } else {
                dp[i][j] = dp[i+1][j] - a[i];
                chmin(dp[i][j], dp[i][j-1] - a[j-1]);
            }
        }
    }

    cout << dp[0][N] << endl;
    return 0;
}