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 <iostream>
#include <cassert>
#include <vector>

using llong = long long;
static const long long INF = 1LL << 60;

template <typename T> std::vector<T> vectors(std::size_t n, T val) {
  return std::vector<T>(n, val);
}

template <typename T, typename... Args>
auto vectors(std::size_t n, Args... args) {
  return std::vector<decltype(vectors<T>(args...))>(n, vectors<T>(args...));
}

template <class T> inline bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

int main()
{
  // Input
  int numArray;
  std::cin >> numArray;
  assert(1 <= numArray and numArray <= 3000);
  std::vector<llong> gain(1 + numArray);
  for (int i = 1; i <= numArray; ++i) {
    std::cin >> gain[i];
    assert(1 <= gain[i] and gain[i] <= 1.0e+9);
  }

  // Initialization
  auto dp = vectors<llong>(1 + numArray, 1 + numArray, -INF);

  // Initial condition
  for (int i = 1; i <= numArray; ++i)
    dp[i][i] = gain[i];

  // DP main
  for (int k = 0; k < numArray; ++k) {
    for (int i = 1; i <= numArray - k; ++i) {
      int j = i + k;
      assert(1 <= i and i <= j and j <= numArray);
      if (j + 1 <= numArray)
        chmax(dp[i][j + 1], gain[j + 1] - dp[i][j]);
      if (i - 1 >= 1)
        chmax(dp[i - 1][j], gain[i - 1] - dp[i][j]);
    }
  }

  // Output
  std::cout << dp[1][numArray] << std::endl;

  return 0;
}