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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <list>
#include <numeric>
#include <stack>
#include <iomanip>
#include <random>
#include <complex>
#include <functional>
#include <tuple>

using namespace std;

#define Rep(i,a,b) for(ll i = a; i < b; ++i)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")

using ll = long long;

constexpr int inf = 1e9 + 7;
constexpr ll infll = 1ll << 60ll;
constexpr ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

namespace {
    template<typename T> void chmax(T& a, T b) { a = std::max(a, b); }
    template<typename T> void chmin(T& a, T b) { a = std::min(a, b); }
    template<typename T> void chadd(T& a, T b) { a = a + b; }

    void Cout(long long x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const T& x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const std::vector<T>& x, const char* sep = " ", const char* end = "\n") { for (std::size_t i = 0, sz = x.size(); i < sz; i++) { std::cout << x[i] << (i == sz - 1 ? end : sep); } }

    // 標準入出力
    struct inp {
        std::size_t sz;
        inp(std::size_t _sz = 1) : sz(_sz) {}
        template <typename T> operator T () const { T a; std::cin >> a; return a; }
        template <typename T> operator std::vector<T>() const { std::vector<T> a(sz); for (std::size_t i = 0; i < sz; i++) std::cin >> a[i]; return a; }
    };

    inp inp1; // input one

    ll rec(int left, int right, int n, const vector<ll>& a, vector<vector<ll>>* dp) {
        if (left >= right) return 0LL;

        int drop_num = (n - right) + left;

        if (right - left == 1) {
            if (drop_num % 2 == 0) return dp->at(left).at(right) = a[left];
            else return dp->at(left).at(right) = -a[left];
        }

        // 右か左か
        ll res = 0LL;
        if (drop_num % 2 == 0) { // 最大化
            res = -infll;
            chmax(res, rec(left + 1, right, n, a, dp) + a[left]);
            chmax(res, rec(left, right - 1, n, a, dp) + a[right - 1]);
        }
        else { // 最小化 
            res = infll;
            chmin(res, rec(left + 1, right, n, a, dp) - a[left]);
            chmin(res, rec(left, right - 1, n, a, dp) - a[right - 1]);
        }

        return dp->at(left).at(right) = res;
    }
}

int main() {

    int n = inp1;
    vector<ll> a = inp(n);

    // dp[i][j] := 左からi個，右からj個取った時の　最適な X-Y
    vector<vector<ll>> dp(n + 1, vector<ll>(n + 1, 0LL));

    Cout(rec(0, n, n, a, &dp));

    // この区間DP，あまり見えないなぁ
    // ゲームAIで活用できそう

    // 階層ごとにどっちが選ぶか分けるとわかりやすい？

    return 0;
}