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.io.*;
import java.lang.*;
import java.util.*;

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

        int N = Integer.parseInt(in.nextLine());
        long[] arr = new long[N];

        StringTokenizer s = new StringTokenizer(in.nextLine());
        for (int i = 0; i < N; ++i) {
            arr[i] = Integer.parseInt(s.nextToken());
        }

        // Stores score relative to Taro
        long[][] dp = new long[N][N];

        for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) dp[i][j] = -1;

        for (int i = 0; i < N; ++i) {
            dp[N-1][i] = arr[i];
        }

        for (int turn = N-2; turn >= 0; --turn) {
            int length = N - turn; // # remaining pieces
            for (int leftMost = 0; leftMost + length <= N; ++leftMost) {

                int l = leftMost;
                int r = leftMost + length - 1;

                dp[turn][leftMost] = Math.max(
                        arr[r] - dp[turn+1][leftMost],
                        arr[l] - dp[turn+1][leftMost+1]);
            }
        }

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

    }
}
