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;

#define int             long long int
#define ld              long double
#define ff              first
#define ss              second
#define mp              make_pair
#define pb              push_back
#define endl            "\n"

#define si              set <int>
#define vi              vector <int>
#define pii             pair <int, int>
#define vpii            vector <pii>
#define mii             map <int, int>
#define que_max         priority_queue <int>
#define que_min         priority_queue <int, vi, greater<int>>

#define loop(i,s,e)     for(int i=s;i<e;i++)
#define rloop(i,e,s)    for(int i=e;i>=s;i--)
#define mset(a,f)       memset(a,f,sizeof(a))


#define M               1000000007
#define sz(x)           ((int) x.size())
#define all(p)          p.begin(), p.end()

#define bug(...)        __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a)        for (auto x : a) cout << x << " "; cout << endl
#define Print(a,x,y)    for (int i=x; i<y; i++) cout<< a[i]<< " "; cout << endl

inline int power(int a, int b)
{
	int x = 1;
	while (b)
	{
		if (b & 1) x *= a;
		a *= a;
		b >>= 1;
	}
	return x;
}

template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
	const char* comma = strchr (names + 1, ',');
	cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}

const int N = 3e5 + 7;
vector <vi > adj;
bool vis[N];
int a[3005];
int dp[3005][3005][3];

int help(int s, int e, int player) {
	if (s > e) return 0;

	if (dp[s][e][player] != -1) return dp[s][e][player];

	if (player == 1) {
		int op1 = a[s] + help(s + 1, e, 2);
		int op2 = a[e] + help(s, e - 1, 2);
		return dp[s][e][player] = max(op1, op2);
	}

	int op1 = help(s + 1, e, 1) - a[s];
	int op2 = help(s, e - 1, 1) - a[e];
	return dp[s][e][player] = min(op1, op2);

}

void solve()
{
	int n;
	cin >> n;
	mset(dp, -1);
	loop(i, 0, n) cin >> a[i];
	cout << help(0, n - 1, 1) ;
	return;

}

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);


	cout << setprecision(9) << fixed;

	int t = 1;
	//cin >> t;
	while (t--)
	{
		solve();
	}


	return 0;
}