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
#include <bits/stdc++.h>
#define ll long long
using namespace std;
long long sum ;
int n;
long long dp[3005][3005],a[3005];
int main(){
	cin>>n;
	for(int i=1;n>=i;i++)
		{
			cin>>a[i];
			sum+=a[i];
		}

	for(int i=1;n>=i;i++)
		dp[i][i]=a[i];
	for(int i=2;n>=i;i++)
	{
		for(int j=1;j+i-1<=n;j++)
		{
			dp[j][j+i-1]=max(a[j]-dp[j+1][j+i-1],-dp[j][j+i-2]+a[i+j-1]);
		}
	}
	cout<<dp[1][n];

	return 0;
}
