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: Accepted)

#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &out, std::vector<T> const &cont) {
    for (auto &i : cont) {
        out << i << ' ';
    }
    return out;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, std::pair<T1, T2> const &p) {
    out << '{' << p.first << ", " << p.second << "}";
    return out;
}
#define imie(...) __debug_print(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __debug_print(const char *name, Arg1 &&arg1) {
#ifdef LOCAL
    cerr << "[" << name << ": " << arg1 << "]" << endl;
#endif
}
template <typename Arg1, typename... Args>
void __debug_print(const char *names, Arg1 &&arg1, Args &&... args) {
#ifdef LOCAL
    const char *comma = strchr(names + 1, ',');
    cerr << "[";
    cerr.write(names, comma - names) << ": " << arg1 << "],  ";
    __debug_print(comma + 2, args...);
#endif
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int N;
    cin >> N;
    int64_t arr[N];
    for(auto& i : arr) cin >> i;
    int64_t DP[N][N];
    for(int l = N-1; l >= 0; --l){
        for(int r = l; r < N; ++r){
            if(l == r){
                DP[l][r] = arr[l];
            }
            else{
                DP[l][r] = max(arr[l] - DP[l+1][r], arr[r] - DP[l][r-1]);
            }
        }
    }
    cout << DP[0][N-1] << endl;
}