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.*;
 
public class Main {

    public static long dp[][] = new long[3001][3001];

    public static long solve(int i,int j,int arr[],int turn){
        if(i > j) return 0;
        if(dp[i][j] != -1){
            return dp[i][j];
        }
        if(turn == 1) dp[i][j] = Math.max( arr[i]+solve(i+1,j,arr,0), arr[j]+solve(i,j-1,arr,0) );
        else dp[i][j] = Math.min(solve(i+1,j,arr,1),solve(i,j-1,arr,1));
        return dp[i][j];
    }

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int arr[] = new int[n];

        long sum = 0;
        for(int i=0;i<n;i++){
            arr[i] = sc.nextInt();
            sum += arr[i];
        }

        for(int i=0;i<=3000;i++){
            for(int j=0;j<=3000;j++){
                dp[i][j] = -1;
            }
        }

        long x = solve(0,n-1,arr,1);
        long y = sum - x;

        System.out.println(x - y);
    }
}