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: Wrong Answer)

#![allow(unused)]

use std::cmp::{min, max};
use std::io::Write;

macro_rules! dbg {
    ($($a:expr),*) => {
        writeln!(
            std::io::stderr(),
            concat!($(stringify!($a), " = {:?}, "),*),
            $($a),*);
    }
}

//
// Based on EbTech's Scanner https://stackoverflow.com/q/56449027
//
struct Scanner<'a, R: std::io::BufRead> {
    reader: R,
    buf: String,
    iter: std::str::SplitWhitespace<'a>,
}

impl<'a, R: std::io::BufRead> Scanner<'a, R> {
    fn new(reader: R) -> Scanner<'a, R> {
        Scanner { reader: reader, buf: String::new(), iter: "".split_whitespace() }
    }

    fn get_str(&mut self) -> &str {
        if let Some(t) = self.iter.next() {
            return t;
        }
        self.buf.clear();
        self.reader.read_line(&mut self.buf).expect("I/O error");
        self.iter = unsafe { std::mem::transmute(self.buf.split_whitespace()) };
        self.get_str()
    }

    fn get<T: std::str::FromStr>(&mut self) -> T {
        self.get_str().parse().ok().expect("Parse error")
    }
}

fn main() {
    let stdin = std::io::stdin();
    solve(Scanner::new(stdin.lock()));
}

fn solve<R: std::io::BufRead>(mut sc: Scanner<R>) {
    let n: usize = sc.get();
    let xs: Vec<i64> = (0..n).map(|_| sc.get()).collect();

    //
    // dp[i][j] = 数列 (a_k) (i + 1 <= k <= j + 1) における X - Y
    //
    let mut dp = vec![vec![0; n]; n];

    dp[0][0] = xs[0];
    dp[n - 1][n - 1] = xs[n - 1];

    for i in (0..(n - 1)).rev() {
        for j in 1..n {
            dp[i][j] = max(xs[i] - dp[i + 1][j], xs[j] - dp[i][j - 1]);
        }
    }

    println!("{}", dp[0][n - 1]);
}