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 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();
		sc.close();
		
		long dp[][] = new long[N+1][N+1];
		for(int i=0;i<N;i++) dp[i][i]=0;
		
		for(int len = 1;len<=N;len++) {
			for(int i=0;i+len<=N;i++) {
				int j=i+len;
				
				//先手番 //N-lenは操作回数を表す。
				if((N-len)%2==0) {
					dp[i][j] = Math.max(dp[i+1][j]+a[i], dp[i][j-1]+a[j-1]);
				}
				else {
					dp[i][j] = Math.min(dp[i+1][j]-a[i], dp[i][j-1]-a[j-1]);
				}
			}
		}
		System.out.println(dp[0][N]);
	}
}
