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)


    #include <bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstring>
    #include <chrono>
    #include <complex>
    #define endl "\n"
    #define ll long long int
    #define vi vector<int>
    #define vll vector<ll>
    #define vvi vector < vi >
    #define pii pair<int,int>
    #define pll pair<long long, long long>
    #define mod 1000000007
    #define inf 1000000000000000001;
    #define all(c) c.begin(),c.end()
    #define mp(x,y) make_pair(x,y)
    #define mem(a,val) memset(a,val,sizeof(a))
    #define eb emplace_back
    #define f first
    #define s second
    
    using namespace std;
    int main()
    {
        std::ios::sync_with_stdio(false);
        int T=1;
        //cin>>T;
        // cin.ignore(); must be there when using getline(cin, s)
        while(T--)
        {
            ll n;
            cin>>n;
            ll a[n];
            for(ll i=0;i<n;i++)
                cin>>a[i];
            ll dp[n+1][n+1];
            for(ll l=n-1;l>=0;l--)
            {
                for(ll r=l;r<n;r++)
                {
                    if(l==r)
                        dp[l][r]=a[l];
                    else
                        dp[l][r]=max(a[l]-dp[l+1][r],a[r]-dp[l][r-1]);
                }
            }
            cout<<dp[0][n-1];
        }
        return 0;
    }