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 (1.15.1) (judge verdict: Accepted)

use std::cmp::max;
use std::error::Error;
use std::io;

fn main() {
    if let Err(_e) = run() {};
}

// it's 1.15 in atcoder, we can't use dyn
fn run() -> Result<(), Box<Error>> {
    let mut buffer = String::new();
    io::stdin().read_line(&mut buffer)?;
    buffer.trim().parse::<usize>()?;

    buffer.clear();
    io::stdin().read_line(&mut buffer)?;

    let inputs = buffer.trim()
        .split_whitespace()
        .map(|a| a.trim().parse::<i64>())
        .fold(Vec::new(), |mut v, a| {
            v.push(a.unwrap());
            v
        });
    let mut dp = vec![vec![-1 as i64; inputs.len()]; inputs.len()];
    println!("{}", calculate(&inputs, &mut dp, 0, inputs.len() - 1));
    return Ok(());
}

fn calculate(inputs: &Vec<i64>, dp: &mut Vec<Vec<i64>>, from: usize, to: usize) -> i64 {
    if dp[from][to] != -1 {
        return dp[from][to];
    }

    if from == to {
        dp[from][to] = inputs[from];
        return inputs[from];
    }
    let first = inputs[from] - calculate(inputs, dp, from + 1, to);
    let last = inputs[to] - calculate(inputs, dp, from, to - 1);

    dp[from][to] = max(first, last);
    return dp[from][to];
}