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 {
    static long[][] dp;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] a = new int[N];
        for(int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
        }
        //dp on the answer for [l,r]
        dp = new long[N][N];
        for(int i = 0; i < N; i++) {
            dp[i][i] = a[i];
        }
        System.out.println(rec(0,N-1,a));
    }
    public static long rec(int l,int r, int[] a){
        if(dp[l][r] > 0) return dp[l][r];
        //l < r guaranteed
        long takeLeft = a[l] - rec(l+1,r,a);
        long takeRight = a[r] - rec(l,r-1,a);
        long ans = Math.max(takeLeft,takeRight);
        dp[l][r] = ans;
        return ans;
    }
}