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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define MOD 1000000007
#define pi 3.141592653589793238
#define ll long long int
#define P pair<ll, ll>
#define VII vector<P>
#define P_B push_back

int n, k;
ll v[3000];
int DP[3000][3000];
ll x = 0, y = 0;
ll backtrack(int beg, int end, bool turn) {

	if(beg > end) return 0;
	
	if(DP[beg][end] != -1) return DP[beg][end];
	ll a, b;
	if(turn) {
		ll a1 = v[beg] + backtrack(beg+1, end, !turn);
		ll a2 = v[end] + backtrack(beg, end-1, !turn);
		return a = max(a1, a2);
	}

	else {
		ll a1 = backtrack(beg+1, end, !turn) - v[beg];
		ll a2 = backtrack(beg, end-1, !turn) - v[end];
		return b = min(a1, a2);
	}

	return DP[beg][end] =  (a-b);
}

int main() {
	
	// ll T; cin>>T; while(T--) 
	{
		
		int n;
		cin>>n;
		memset(DP, -1, sizeof(DP));
		for (int i = 0; i < n; i++) cin>>v[i];
		cout<<backtrack(0, n-1, true);

	}	
	return 0;
}