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;

void getResult(int a, int b, long long* vals, long long** result) {

	if (result[a][b] != -1) {
		return;
	}

	if (a == b) {
		result[a][b] = vals[a];
		return;
	}

	getResult(a + 1, b, vals, result);
	getResult(a, b - 1, vals, result);

	if (result[a + 1][b] < result[a][b - 1]) {
		result[a][b] = result[a + 1][b] + vals[a];
	}

	else {

	}

	result[a][b] = max(vals[a] - result[a + 1][b], vals[b] - result[a][b - 1]);
	return;

}

int main() {

	int n;
	cin >> n;

	long long vals[n];
	for (int i = 0; i < n; i++) {
		cin >> vals[i];
	}

	// Resulting value of x-y for a game only including values from indices a to b
	long long** result = new long long*[n];
	for (int i = 0; i < n; i++) {
		result[i] = new long long[n];
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			result[i][j] = -1l;
		}
	}

	getResult(0, n-1, vals, result);
	cout << result[0][n-1] << endl;

}