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.*;
import java.io.*;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String[] s = br.readLine().split(" ");
		int[] arr = new int[n + 1];
		long[] rsum = new long[n + 1];
		long[][] dp = new long[n + 1][n + 1];
		
		for (int i = 1; i <= n; i++) {
			arr[i] = Integer.parseInt(s[i - 1]);
			dp[i][i] = arr[i];
		}			

		for (int i = 2; i <= n; i++) {
			for (int j = i - 1; j > 0; j--) {
				dp[i][j] = Math.max(arr[j] - dp[i][j+1], arr[i] - dp[i-1][j]);
				
//				System.out.print("[" + i + " " + j + "](" + dp[i][j][0] + " " + dp[i][j][1] + ") ");
			}
//			System.out.println();
		}

		System.out.println(dp[n][1]);
	}

}

//use for actual values OR use system of linear equation
//long first = dp[i][j + 1][1] + arr[j];
//long last = dp[i - 1][j][1] + arr[i];
//
//if (first > last) {
//	dp[i][j][0] = first;
//	dp[i][j][1] = rsum[i] - rsum[j - 1] - first;
//} else {
//	dp[i][j][0] = last;
//	dp[i][j][1] = rsum[i] - rsum[j - 1] - last;
//}
