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;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(in);
        int n = scanner.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scanner.nextInt();
        }
        long[][] dp = new long[n][n];
        long s = sum(arr,0,n-1,dp);

        System.out.println(s);
    }
    private static long sum(int[] arr, int i, int j, long[][] dp){
        if(i>j) return 0;
        if(i==j) return arr[i];
        if(dp[i][j]!=0) return  dp[i][j];
        return dp[i][j] = Math.max(arr[i]-sum(arr,i+1,j,dp ),arr[j]-sum(arr,i,j-1,dp ));
    }
}