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;
#define int long long
#define pb              push_back
#define mp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define mod 			1000000007
#define tc(t) 			int t;cin >> t;while(t--)
#define for0(i , n)		for(int i=0;i<n;i++)
#define loop(i , a, b)	for(int i=a;i<=b;i++)
#define endl 			'\n'
#define inf 			1e18
#define fi 				first
#define se 				second

int XX[8] = { +1, +1, +1, 0, 0, -1, -1, -1};
int YY[8] = { +1, 0, -1, +1, -1, +1, 0, -1};
int dx[4] = { -1, 1, 0, 0};
int dy[4] = {0, 0, 1, -1};

int n;
int a[3001];
int dp[3001][3001][2];

int solve(int i, int j, int fl) {
	if (i > j) return 0;
	if (dp[i][j][fl] != -1) return dp[i][j][fl];
	if (fl) {
		return dp[i][j][fl] = max(a[i] + solve(i + 1, j, 0), a[j] + solve(i, j - 1, 0));
	}

	return dp[i][j][fl] = min(solve(i + 1, j, 1), solve(i, j - 1, 1));
}

signed main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

#ifndef ONLINE_JUDGE
	freopen("input.txt" , "r" , stdin);
	freopen("output.txt" , "w" , stdout);
#endif

	cin >> n;
	int sum = 0;
	for0(i, n) cin >> a[i], sum += a[i];

	memset(dp, -1, sizeof(dp));

	int x = solve(0, n - 1, 1);
	int y = sum - x;

	cout << x - y << endl;

	return 0;
}