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.PrintWriter;
public class Main{
	static long[] a;
	static long[][] dp;
	static boolean[][] visited;
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		a=new long[n];
		dp=new long[n][n];
		visited=new boolean[n][n];
		for(int i=0; i<n; i++){
			a[i]=sc.nextLong();
		}
		System.out.println(dfs(0,n-1));
	}
	static long dfs(int l,int r){
		if(visited[l][r]) return dp[l][r];
		if(l==r) return a[l];
		visited[l][r]=true;
		return dp[l][r]=Math.max(a[l]-dfs(l+1,r),a[r]-dfs(l,r-1));
	}
}