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)

#[allow(unused_macros)]
macro_rules! input {
    ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; let mut next = || { iter . next ( ) . unwrap ( ) } ; input_inner ! { next , $ ( $ r ) * } };
    ( $ ( $ r : tt ) * ) => { let stdin = std :: io :: stdin ( ) ; let mut bytes = std :: io :: Read :: bytes ( std :: io :: BufReader :: new ( stdin . lock ( ) ) ) ; let mut next = move || -> String { bytes . by_ref ( ) . map ( | r | r . unwrap ( ) as char ) . skip_while ( | c | c . is_whitespace ( ) ) . take_while ( | c |! c . is_whitespace ( ) ) . collect ( ) } ; input_inner ! { next , $ ( $ r ) * } };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ( $ next : expr ) => { };
    ( $ next : expr , ) => { };
    ( $ next : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ next , $ t ) ; input_inner ! { $ next $ ( $ r ) * } };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ( $ next : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ next , $ t ) ) ,* ) };
    ( $ next : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ next , $ t ) ) . collect ::< Vec < _ >> ( ) };
    ( $ next : expr , chars ) => { read_value ! ( $ next , String ) . chars ( ) . collect ::< Vec < char >> ( ) };
    ( $ next : expr , usize1 ) => { read_value ! ( $ next , usize ) - 1 };
    ( $ next : expr , $ t : ty ) => { $ next ( ) . parse ::<$ t > ( ) . expect ( "Parse error" ) };
}

#[allow(unused_macros)]
macro_rules! vec_md {
    ( $ init : expr ; $ ( $ dim : expr ) ;* ) => { vec_md ! ( $ ( $ dim ) ;* => $ init ) };
    ( $ head : expr ; $ ( $ tail : expr ) ;* => $ init : expr ) => { vec ! [ vec_md ! ( $ ( $ tail ) ;* => $ init ) ; $ head ] };
    ( $ head : expr => $ init : expr ) => { vec ! [ $ init ; $ head ] };
}

fn main() {
    input! {
        n: usize,
        a: [i64; n]
    }

    let mut dp = vec_md![0i64; n+1; n+1];
    for len in 1..n+1 {
        for i in (0..).take_while(|&i| i + len <= n) {
            let j = i + len;
            if (n - len) % 2 == 0 {
                dp[i][j] = std::cmp::max(
                    dp[i+1][j] + a[i],
                    dp[i][j-1] + a[j-1]
                );
            } else {
                dp[i][j] = std::cmp::min(
                    dp[i+1][j] - a[i],
                    dp[i][j-1] - a[j-1]
                );
            }
        }
    }

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