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

public class Main 
{   
    static long[] A;
    static long[][] dp;
    
    static long fn(int i, int j, int c) {
        if (i > j) { return 0; }
        if (dp[i][j] != -1) { return dp[i][j]; }
        if (c == 0) {
            dp[i][j] = Math.max(A[i] + fn(i+1, j, 1), A[j] + fn(i, j-1, 1));
        }
        else {
            dp[i][j] = Math.min(-A[i] + fn(i+1, j, 0), -A[j] + fn(i, j-1, 0));
        }
        return dp[i][j];
    }
    
    public static void main(String[] args) throws IOException 
    {
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        A = new long[N];
        for (int n =0; n < N; n ++) {
            A[n] = scan.nextLong();
        }
        dp = new long[N][N];
        for (int i =0; i < N; i ++) {
            for (int j = 0; j < N; j ++) {
                dp[i][j] = -1;
            }
        }
        System.out.println(fn(0, N-1, 0));
    }
}    
