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.Arrays;
import java.util.Scanner;

public class Main {

	static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
		int N = sc.nextInt();
		long[] A = new long[N];
		for (int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
		}
		long[][] dp = new long[N + 1][N + 1];
		for (int l = 1; l <= N; l++) {
			for (int i = 0; i + l <= N; i++) {
				int j = i + l;
				boolean first = (N - l) % 2 == 0;
				if (first) {
					dp[i][j] = dp[i + 1][j] + A[i];
					dp[i][j] = Math.max(dp[i][j], dp[i][j - 1] + A[j - 1]);
				} else {
					dp[i][j] = dp[i + 1][j] - A[i];
					dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] - A[j - 1]);
				}
			}
		}
		System.out.println(dp[0][N]);
	}
}