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;
long long int Mod=1e9+7;
long long  dp[3005][3005];
long long rec(long long A[],long i,long j)
{
    if(i>j)
        return(0);
    else if(i==j)
    {
        dp[i][i]=A[i];
        return(dp[i][i]);
    }
    else if(i==j-1)
    {
        dp[i][i+1]=max(A[i],A[i+1]);
        return(dp[i][i+1]);
    }
    else if(i==j-2)
    {
        dp[i][i+2]=max(A[i]+min(A[i+1],A[i+2]),A[i+2]+min(A[i],A[i+1]));
        return(dp[i][i+2]);
    }
    if(dp[i][j]!=-1)
        return(dp[i][j]);
    else
    {
    dp[i][j]=max(min(rec(A,i+2,j),rec(A,i+1,j-1))+A[i],min(rec(A,i+1,j-1),rec(A,i,j-2))+A[j]);
        return(dp[i][j]);
        
    }
}
int main()
{
    long i,j,n,k,ans,sum=0;
    cin>>n;
    long long A[n+1];
    A[0]=-1;
    for(i=1;i<=n;i++)
    {
        cin>>A[i];
        sum+=A[i];
    }
    
    for(i=0;i<=n;i++)
        for(j=0;j<=n;j++)
            dp[i][j]=-1;
    
    ans = rec(A,1,n);
    cout<<2*ans-sum<<endl;
}