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();
	
	int arr[]=new int[n];
	long sum=0;
	for(int i=0;i<n;i++) {
		arr[i]=s.nextInt();
		sum+=arr[i];
	}
	long dp[][]=new long[n+1][n+1];
	for(int i=0;i<=n;i++) {
		for(int j=0;j<=n;j++) {
			dp[i][j]=-1;
		}
	}
	long Xans=f(0,n-1,n,arr,dp);
	long Yans=sum-Xans;
	 System.out.print(Xans-Yans);
	 
   }
    public static long f(int start,int end,int n,int arr[],long dp[][]) {
    	if(start>end) {
    		return 0;
    	}
    	if(dp[start][end]!=-1) {
    		return dp[start][end];
    	}
    	long a=arr[start]+ Math.min(f(start+2,end,n,arr,dp),
    			f(start+1,end-1,n,arr,dp));
    	long b=arr[end]+ Math.min(f(start,end-2,n,arr,dp),
    			f(start+1,end-1,n,arr,dp));
    	return dp[start][end]=(a>b)?a:b;
    }
}