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>
using namespace std;
typedef long long ll;
typedef long double ld;
#define int ll
//#define double ld
#define faster ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
ifstream in("input.txt");ofstream out("output.txt");
const int inf=1e18;
const int mod=1e9+7;
int n,k;
vector<int> vec;
vector<vector<int> > dp;

void init()
{
    cin>>n;
    vec.resize(n);
    for(int i=0;i<n;i++)
        cin>>vec[i];
    dp.resize(n,vector<int>(n,inf));
    for(int i=0;i<n;i++)
    {
        dp[i][i]=0;
        if((i+i)%2==(n-1)%2)
            dp[i][i]+=vec[i];
        else
            dp[i][i]-=vec[i];
    }
}

int go(int l,int r)
{
    if(l<0||r<0||l>=n||r>=n)
        return inf;
    if(dp[l][r]!=inf)
        return dp[l][r];
    int ans;
    if((l+r)%2==(n-1)%2)
    {
        ans=-inf;
        if(go(l+1,r)!=inf)
            ans=max(ans,go(l+1,r)+vec[l]);
        if(go(l,r-1)!=inf)
            ans=max(ans,go(l,r-1)+vec[r]);
    }
    else
    {
        ans=min(-vec[l]+go(l+1,r),-vec[r]+go(l,r-1));
    }
    return dp[l][r]=ans;
}

signed main()
{
    init();
    cout<<go(0,n-1)<<"\n";
    return 0;
}