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 <iostream>
#include <array>
#include <algorithm>
#include <vector>
#include <bitset>
#include <set>
#include <unordered_set>
#include <cmath>
#include <complex>
#include <deque>
#include <iterator>
#include <numeric>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <limits>
#include <iomanip>
#include <functional>
#include <cassert>
using namespace std;

using ll=long long;
template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
#define rep(i, k, n) for (ll i=k; i<(ll)n; ++i)
#define REP(i, n) rep(i, 0, n)
#define ALL(v) v.begin(),v.end()
template < class T > inline bool chmax(T& a, T b) {if (a < b) { a=b; return true; } return false; }
template < class T > inline bool chmin(T& a, T b) {if (a > b) { a=b; return true; } return false; }
#define DEBUG_VLL(vec) REP(sz, vec.size()) std::cerr<<vec[sz]<<(sz==vec.size()-1?'\n':' ');

const long long MOD = 1000000007;
const long long HIGHINF = (long long)1e18;
const int INF = (int)1e9;

vll a;
vvll dp;

ll dfs(int left, int right) {
    if (dp[left][right] > -HIGHINF) return dp[left][right];
    if (left == right) return dp[left][right] = a[left];
    return dp[left][right] = max(a[left] - dfs(left + 1, right), a[right] - dfs(left, right - 1));
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    a.resize(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    dp.resize(n, vll(n, -HIGHINF));
    cout << dfs(0, n - 1) << '\n';
    return 0;
}