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>

#define MAXN 3000

long S[2][MAXN][MAXN];
int T[2][MAXN][MAXN];
int A[MAXN];

long deque ( int n, int m, int p)
{
	if ( T[p][n][m] == 1 ) return S[p][n][m];

	long tmp1,tmp2;
	if ( p == 0 ) {
		tmp1 = deque(n+1, m, 1)+A[n];
		tmp2 = deque(n,m-1,1)+A[m];
		T[p][n][m] = 1;
		return S[p][n][m] = (tmp1 > tmp2 ) ? tmp1:tmp2;
	} else {
		tmp1 = deque(n+1,m,0)-A[n];
		tmp2 = deque(n,m-1,0)-A[m];
		T[p][n][m] = 1;
		return S[p][n][m] = (tmp1 < tmp2 ) ? tmp1:tmp2;
	}
}

int main(void)
{
	int n,i,j;

	scanf("%d", &n);
	for ( i = 0; i < n; i++ ) scanf("%d", &A[i]);
	for ( i = 0; i < n; i++ ) {
		for ( j = 0; j < n; j++ ) T[0][i][j] = T[1][i][j] = 0;
	}
	for ( i = 0; i < n; i++ ) {
		S[0][i][i] = A[i], T[0][i][i] = 1;
		S[1][i][i] = -A[i], T[1][i][i] = 1;
	}

	printf("%ld\n",deque(0,n-1,0));

	return 0;
}