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 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int count = scanner.nextInt();
        int[] arr = new int[count];
        for (int i = 0; i < count; ++i) {
            arr[i] = scanner.nextInt();
        }

        long[][] dp = new long[count + 1][count + 1];
        for (int piles = 1; piles <= count; ++piles) {
            for (int start = 0; start + piles <= count; ++start) {
                if ((count - piles) % 2 == 0) {
                    dp[piles][start] = Math.max(
                            dp[piles - 1][start + 1] + arr[start],
                            dp[piles - 1][start] + arr[start + piles - 1]);
                } else {
                    dp[piles][start] = Math.min(
                            dp[piles - 1][start + 1] - arr[start],
                            dp[piles - 1][start] - arr[start + piles - 1]);
                }
            }
        }

        System.out.println(dp[count][0]);
    }
}
