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)

//g++  5.4.0

#include<bits/stdc++.h>
using namespace std;

#define ff first
#define ss second
#define int long long
#define ld long double
#define pb push_back
#define pii pair< int,int >
#define fast ios::sync_with_stdio(0) , cin.tie(0) , cout.tie(0) ;



signed main()
{
    fast;
    int n;
    cin >> n ;
    vector<int> a(n+1,0) ;
    for(int i=1 ; i<=n ; i++ )
    {
        cin >> a[i];
        a[i] += a[i-1];
    }
    int dp[n+1][n+1];
    memset( dp,0,sizeof(dp) );
    
    for(int i=1 ; i<=n ; i++ )
        dp[i][i] = a[i]-a[i-1];
    
    for(int len=2 ; len<=n ; len++ )
    {
        for(int st=1 ; st+len-1<=n ; st++ )
        {
            int i = st;
            int j = st+len-1;
            dp[i][j] = max( -a[i-1] + a[j] - dp[i+1][j] , a[j] - a[i-1] - dp[i][j-1] );
        }
    }
    //cout << dp[1][n] << "\n";
    cout << dp[1][n] - ( a[n]-dp[1][n] ) << "\n";
}