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 int n;
	static int[] a;
	static long[][][] dp;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		sc.close();

		dp = new long[2][n][n];
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < n; j++) {
				Arrays.fill(dp[i][j], Long.MIN_VALUE);
			}
		}
		System.out.println(dfs(0, 0, n - 1));
	}

	static long dfs(int t, int l, int r) {
		if (dp[t][l][r] != Long.MIN_VALUE) {
			return dp[t][l][r];
		}
		if (t == 0) {
			if (l == r) {
				return dp[t][l][r] = a[l];
			}
			long ret = Long.MIN_VALUE;
			ret = Math.max(ret, dfs(1, l + 1, r) + a[l]);
			ret = Math.max(ret, dfs(1, l, r - 1) + a[r]);
			return dp[t][l][r] = ret;
		} else {
			if (l == r) {
				return dp[t][l][r] = -a[l];
			}
			long ret = Long.MAX_VALUE;
			ret = Math.min(ret, dfs(0, l + 1, r) - a[l]);
			ret = Math.min(ret, dfs(0, l, r - 1) - a[r]);
			return dp[t][l][r] = ret;
		}
	}
}
