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 <class T> ostream& operator << (ostream &os, const vector<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T> ostream& operator << (ostream &os, const set<T> &v) { for (T i : v) os << i << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const pair<T, S> &v) { os << v.first << ' ' << v.second; return os; }
template <class T, class S> ostream& operator << (ostream &os, const map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
template <class T, class S> ostream& operator << (ostream &os, const unordered_map<T, S> &v) { for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' '; return os; }
 
#ifndef ONLINE_JUDGE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <class Arg1> void __f(const char* name, Arg1&& arg1) { cerr << name << " : " << arg1 << endl; }
    template <class Arg1, class... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args) {
        const char* sep = strchr(names + 1, ',');
        cerr.write(names, sep - names) << " : " << arg1 << "  ";
        __f(sep + 1, args...);
    }
#else
#define trace(...) 0
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#endif
 
#define fast_cin() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define MAX 1000000
#define ll long long
#define int ll
#define db double
#define str string
#define pb push_back
#define For(i,s,e) for (ll i=(s); i<(e); i++)
#define Forrev(i,s,e) for (ll i=(s); i>=(e); i--)
#define all(v) v.begin(),v.end()
 
#define vll vector<ll>
#define vs vector<string>
#define mapll map<ll,ll>
#define pll pair<ll,ll>
#define initialise(a, x) memset(a, x, sizeof(a))
#define maxheap priority_queue<ll>
#define minheap priority_queue<ll,vector<ll> ,greater<ll>>
 
#define ff first
#define ss second
#define endl "\n"
#define mp make_pair
const ll mod=1e9 + 7;
 
ll takemod(ll a) {
    return ((a%mod)+mod)%mod;
}
 
ll gcd(ll a, ll b) { 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);        
}
 
ll fast_exp(ll base, ll expo) {
    ll res=1;
    while(expo>0) {
        if(expo&1) res=(res*base)%mod;
        base=(base*base)%mod;
        expo>>=1;}
    return res;
}
 
ll modinv(ll a) {
    return takemod(fast_exp(takemod(a), mod-2));
}
 
void solve() {
	int n;
	cin >> n;

	int arr[n];
	for (int i = 0; i < n; i++)
		cin >> arr[i];
	
	int dp[n][n][2];
	
	for (int i = 0; i < n; i++) {
		dp[i][i][0] = arr[i];
		dp[i][i][1] = -arr[i]; 
	}

	for (int i = 1; i <n; i++) {
		
		for (int j = 0; j+i < n; j++) {
			dp[j][j+i][0] = max(dp[j+1][j+i][1] + arr[j], dp[j][i+j-1][1] + arr[j+i]);
			dp[j][j+i][1] = min(dp[j+1][j+i][0] - arr[j], dp[j][i+j-1][0] - arr[j+i]); 
		}
	}
	
	cout << dp[0][n-1][0] << endl;

}

 
signed main() 
{
    fast_cin();
    
    
    int t=1;
    // cin>>t;
    while(t--)
    {
        solve();
    }
    
    
    return 0;    
}