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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use text_io::*;

fn main() {
    const U_INF: usize = 1 << 60;
    const I_INF: isize = 1 << 60;

    let n: usize = read!();
    let mut a = Vec::with_capacity(n);
    for _ in 0..n {
        let ai: isize = read!();
        a.push(ai);
    }

    let mut dp = vec![vec![None; n + 10]; n + 10];
    for i in 0..=n {
        dp[i][i] = Some(0);
    }

    // for len in 1..=n {
    //     for i in 0..n {
    //         let j = len + i;
    //         if j <= n {
    //             if (n - len) % 2 == 0 {
    //                 let jj = if j == n { -I_INF } else { dp[i][j - 1] + a[j] };
    //                 dp[i][j] = (dp[i + 1][j] + a[i]).max(jj);
    //             } else {
    //                 let jj = if j == n { I_INF } else { dp[i][j - 1] - a[j] };
    //                 dp[i][j] = (dp[i + 1][j] - a[i]).min(jj);
    //             }
    //         }
    //     }
    // }

    println!("{}", memo(n, 0, n, &mut dp, &a));
}

fn memo(n: usize, i: usize, j: usize, dp: &mut Vec<Vec<Option<isize>>>, a: &Vec<isize>) -> isize {
    match dp[i][j] {
        Some(ans) => ans,
        None => {
            if (n - (j - i)) % 2 == 0 {
                let ans =
                    (memo(n, i + 1, j, dp, a) + a[i]).max(memo(n, i, j - 1, dp, a) + a[j - 1]);
                dp[i][j] = Some(ans);
                ans
            } else {
                let ans =
                    (memo(n, i + 1, j, dp, a) - a[i]).min(memo(n, i, j - 1, dp, a) - a[j - 1]);
                dp[i][j] = Some(ans);
                ans
            }
        }
    }
}