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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

uses math;
const fi='l_deque.inp';
      fo='l_deque.out';
      maxn=3000;
var n:int32;
    sum,f:array[0..maxn,0..maxn] of int64;
    a:array[1..maxn] of int32;
//=============================
procedure nhap;
var i:int32;
begin
   readln(n);
   for i:=1 to n do
   begin read(a[i]); sum[1,i]:=sum[1,i-1]+a[i]; end;
end;
//=============================
function max(x,y:int64):int64;
begin
   if x>y then exit(x)
   else exit(y);
end;
//=============================
procedure xuli;
var i,j:int32;
begin
   for i:=1 to n do
    for j:=n downto i do sum[i,j]:=sum[1,j]-sum[1,i-1];
   for i:=1 to n do f[i,i]:=a[i];
   for i:=n-1 downto 1 do
    for j:=i+1 to n do
     f[i,j]:=max(sum[i,j]-f[i+1,j],sum[i,j]-f[i,j-1]);
   write(2*f[1,n]-sum[1,n]);
end;
//=============================
begin
//   assign(input,fi); reset(input);
//   assign(output,fo); rewrite(output);
   nhap;
   xuli;
//   close(input); close(output);
end.