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>

using namespace std;


int main() {
	int N;
    cin >> N;

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

    // 区間[a_l, a_r)が残っている場合での最適な行動をとった場合のX-Yの最大値dp[l][r]
    vector<vector<long long>> dp(N+1, vector<long long>(N+1, 0));

    // 初期条件
    for (int l = 0; l < N; l++) {
        dp[l][l] = 0;
    }

    // dp開始
    for (int len = 1; len <= N; len++) {
        for (int l = 0; l + len <= N; l++) {
            int r = l + len;

            // 先手
            if ((N-len) % 2 == 0) {
                dp[l][r] = max(dp[l+1][r] + a[l], dp[l][r-1] + a[r-1]);
            } else { //後手
                dp[l][r] = min(dp[l+1][r] - a[l], dp[l][r-1] - a[r-1]);
            }
        }
    }

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