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 void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		long[] arr = new long[n];
		for (int i = 0; i < n; i++) {
			arr[i] = s.nextLong();
		}
		long[][] dp = new long[n][n];
		for (int i = 0; i < n; i++)
			Arrays.fill(dp[i], Long.MIN_VALUE);
		System.out.println(rec(arr, 0, n - 1, dp));
	}

	public static long rec(long[] arr, int i, int j, long[][] dp) {
		if (i == j) {
			dp[i][j] = arr[i];
			return dp[i][j];
		}
		if (dp[i][j] != Long.MIN_VALUE)
			return dp[i][j];

		dp[i][j] = Math.max(arr[i] - rec(arr, i + 1, j, dp), arr[j] - rec(arr, i, j - 1, dp));
		return dp[i][j];
	}
}
