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;
using pii=pair<int,int>;
using ll=long long;

#define fi first
#define se second
#define pb push_back
#define rep(i,n) for(int i=0; i<(int)(n); i++)

ll dp[3010][3010];
ll a[3010];

ll rec(int i, int j, bool b) {
	ll res;
	if(dp[i][j] != -1e18){
		return dp[i][j];
	} 
	else {
		if(b){
			 res = max(rec(i,j-1,false)+a[j], rec(i+1,j,false) + a[i]);
		} 
		else {
			 res = min(rec(i+1,j,true)-a[i], rec(i,j-1,true)-a[j]);
		}
	}
	return dp[i][j] = res;
}

int main() {
	int n;
	cin >> n;
	rep(i,n) {
		cin >> a[i];
	}
	rep(i,n) rep(j,n) dp[i][j] = -1e18;
	rep(i,n) {
		if(n%2==1) dp[i][i] = a[i];
		else dp[i][i] = -1*a[i];
	}
	cout << rec(0,n-1,true) << endl;
	/*rep(i,n) {
		rep(j,n) {
			if(dp[i][j] == -1e18) cout << setw(3) << "NA";
			else cout << setw(3) << dp[i][j];
		}
		cout << endl;
	}*/

  return 0;
}