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>

int N;
int a[3000];
long long dp[3000][3000];

long long dfs(int l, int r){
  if(dp[l][r] != -1) return dp[l][r];
  if(l == r) return dp[l][r] = ((N & 1) ? a[l] : -a[l]);
  
  long long left, right;
  if((N - (r - l)) & 1){
    left = dfs(l + 1, r) + a[l];
    right = dfs(l, r - 1) + a[r];
    if(left > right) dp[l][r] = left;
    else dp[l][r] = right;
  }else{
    left = dfs(l + 1, r) - a[l];
    right = dfs(l, r - 1) - a[r];
    if(left < right) dp[l][r] = left;
    else dp[l][r] = right;
  }
  
  return dp[l][r];
}

int main(){
  scanf("%d", &N);
  
  for(int i = 0; i < N; i++) scanf("%d", &a[i]);
  
  for(int l = 0; l < N; l++){
    for(int r = l; r < N; r++){
      dp[l][r] = -1;
    }
  }
  
  printf("%lld\n", dfs(0, N - 1));
  
  return 0;
}