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.

Examples

Input

4
10 80 90 30


Output

10


Input

3
10 100 10


Output

-80


Input

1
10


Output

10


Input

10
1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000 1


Output

4999999995


Input

6
4 2 9 7 1 5


Output

2
import java.util.Scanner;

public class Main {
    static Long[][] dp;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int N = scanner.nextInt();
        int[] stones = new int[N];

        dp = new Long[N + 1][N + 1];

        for (int i = 0; i < stones.length; i++) {
            stones[i] = scanner.nextInt();
        }

        System.out.printf("%d\n", recurse(stones, 0, stones.length - 1));
    }

    private static long recurse(int[] stones, int l, int r) {
        if (l == r) {
            return stones[l];
        }

        if (dp[l][r] != null) {
            return dp[l][r];
        }

        dp[l][r] = Math.max(stones[l] - recurse(stones, l + 1, r), stones[r] - recurse(stones, l, r - 1));
        return dp[l][r];
    }
}