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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdbool.h>

int n;
long long int a[3000];
long long int dp[3001][3002][2];
bool check[3001][3002][2];
long long int f();
long long int max(long long int a, long long int b){
    return (a >= b) ? a : b;
}
long long int min(long long int a, long long int b){
    return (a <= b) ? a : b;
}

int main(int argc, char const *argv[])
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
    }
    for (int i = 0; i < 3001; i++)
    {
        for (int j = 0; j < 3002; j++)
        {
            for (int k = 0; k < 2; k++)
            {
                dp[i][j][k] = 0;
                check[i][j][k] = (i == j) ? true : false;
            }
            
        }
        
    }
    printf("%lld\n", f(0, n, 0));

    return 0;
}

long long int f(int left, int right, int tern)
{
    long long int tmp;    
    if (check[left][right][tern])
        return dp[left][right][tern];
    check[left][right][tern] = true;
    if(tern == 0){
        tmp = max(f(left + 1, right, tern ^ 1) + a[left], f(left, right - 1, tern ^ 1) + a[right - 1]);
    } else {

        tmp = min(f(left + 1, right, tern ^ 1) - a[left], f(left, right - 1, tern ^ 1) - a[right - 1]);
        
    }
    dp[left][right][tern] = tmp;
    return tmp;
}