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)

#define _CRT_SECURE_NO_WARNINGS
#define _SCL_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <utility>
#include <algorithm>
#include <functional>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <limits>
#include <numeric>
#include <valarray>
#include <fstream>

using namespace std;
typedef unsigned int uint;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, LL> PP;
#define REP(i, a, n) for(LL i = (a), i##_max = (n); i < i##_max; ++i)
#define REM(i, a, n) for(LL i = (LL)(n) - 1, i##min = (a); i >= i##min; --i)
#define ALL(arr) (arr).begin(), (arr).end()
#define FLOAT fixed << setprecision(16)
#define SPEEDUP {cin.tie(NULL); ios::sync_with_stdio(false);}
const int INF = 0x3FFFFFFF;
const LL INFLL = 0x3FFFFFFF3FFFFFFF;
const double INFD = 1.0e+308;
const string INFSTR = "\x7f";
const double EPS = 1.0e-9;

void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
template <class T, class U>
istream& operator>>(istream& ist, pair<T, U>& right) { return ist >> right.first >> right.second; }
template <class T, class U>
ostream& operator<<(ostream& ost, const pair<T, U>& right) { return ost << right.first << ' ' << right.second; }
template <class T, class TCompatible, size_t N>
void Fill(T(&dest)[N], const TCompatible& val) { fill(dest, dest + N, val); }
template <class T, class TCompatible, size_t M, size_t N>
void Fill(T(&dest)[M][N], const TCompatible& val) { for (int i = 0; i < M; ++i) Fill(dest[i], val); }
template<class T>
T Compare(T left, T right) { return left > right ? 1 : (left < right ? -1 : 0); }
istream& Ignore(istream& ist) { string s; ist >> s; return ist; }
bool Inside(int i, int j, int h, int w) { return i >= 0 && i < h && j >= 0 && j < w; }
template <class T>
T Next() { T buf; cin >> buf; return buf; }

#ifdef ONLY_MY_ENVIR
#include "IntMod.h"
#include "UnionFind.h"
#include "Graph.h"
#include "Range.h"
#include "Global.h"
#include "FlowSolver.h"
#include "Tree.h"
#include "SuffixArray.h"
#include "Geometry.h"
#include "Matrix.h"
#include "SegmentTree.h"
#include "SegmentTree2D.h"
#include "BIT.h"
#include "Rational.h"
#include "Position.h"
#include "Factorization.h"
#include "Math.h"
#endif

#ifdef __GNUC__
typedef __int128 LLL;
istream& operator>> (istream& ist, __int128& val) { LL tmp;  ist >> tmp; val = tmp; return ist; }
ostream& operator<< (ostream& ost, __int128 val) { LL tmp = val; ost << tmp; return ost; }
#endif

#if 1234567891
#include <array>
#include <random>
#include <unordered_set>
#include <unordered_map>
template<typename T>
using PriorityQ = priority_queue<T, vector<T>, greater<T> >;
// template <class T>
// auto Is(const T& value) { return [value](const auto& comparand) -> bool { return comparand == value; }; }
#endif

int N;
int A[3000];
LL DP[3001][3001];
int main() {
	cin >> N;
	REP(i, 0, N) {
		cin >> A[i];
	}

	REP(t, 1, N + 1) {
		REP(i, 0, N) {
			int j = i + t;
			if (j > N) continue;
			
			bool taro = (N - t) % 2 == 0;

			if (taro) {
				DP[i][j] = max(
					DP[i + 1][j] + A[i],
					DP[i][j - 1] + A[j - 1]
				);
			} else {
				DP[i][j] = min(
					DP[i + 1][j] - A[i],
					DP[i][j - 1] - A[j - 1]
				);
			}
		}
	}

	cout << DP[0][N] << endl;
	return 0;
}