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;
#define forn(i,n) for(int i=0;i<n;i++)
#define ALL(x) x.begin(),x.end()
#define LL long long int

int mod = pow(10,9) +7;
const int inf = 2e9;
const LL linf = 2e18;
const double eps = 1e-9;

/////////////////////////////////////////////

const int N = 3e3+100;
LL arr[N];
LL dp[N][N];

LL total(LL x,LL y , bool c ){
    // trace(x,y,c);
    if(dp[x][y] != -1LL) return dp[x][y] ;
    if(x > y ) return 0;
    if(c == 0) return  dp[x][y ]=  max( arr[x] + total(x+1,y,!c) , arr[y] + total(x,y-1,!c) );
    else return dp[x][y] =  min(total(x+1,y,!c) -arr[x],total(x,y-1,!c) -arr[y]);
} 


int solve(int x, int y,int c){
    if(x == y){
        return arr[x];
    }
    int mini = min(arr[x] + solve(x+1,y,c^1),arr[y] + solve(x,y-1,c^1));
    int maxi = min(arr[x] + solve(x+1,y,c^1),arr[y] + solve(x,y-1,c^1));

    int play1 = min( arr[x] + solve(x+1,y,1) - solve(x+1,y,0) ,arr[y] +  solve(x,y-1,1) - solve(x,y-1,0) );
    int play2 = min( arr[x] + solve(x+1,y,1) - solve(x+1,y,0) ,arr[y] +  solve(x,y-1,1) - solve(x,y-1,0) );
    if(c == 0){
        return play1;
    }else{
        return play2;
    }
}


int main()
{
    
    LL n;
    cin >> n;
    forn(i,n) forn(j,n) dp[i][j] = -1LL;
    forn(i,n) cin >> arr[i];
    cout << total(0,n-1,0) << endl;
}