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
# -*- coding: utf-8 -*-
import sys

N=input()
A=map(int, sys.stdin.readline().split())

dp=[ [ None for i in range(N) ]  for j in range(N) ]


for i in range(N-1,-1,-1):
	if N%2==0:
		dp[i][i]=A[i]*-1
	else:
		dp[i][i]=A[i]


for i in range(N-2,-1,-1):
	for j in range(i+1,N):
		t=N-(j-i+1)	#残りの要素数
		if t%2==0:
			if dp[i][j] is None : dp[i][j]=float("-inf")
			dp[i][j]=max(dp[i][j] , dp[i][j-1]+A[j])
			dp[i][j]=max(dp[i][j] , dp[i+1][j]+A[i])
		else:
			if dp[i][j] is None : dp[i][j]=float("inf")
			dp[i][j]=min(dp[i][j] , dp[i][j-1]-A[j])
			dp[i][j]=min(dp[i][j] , dp[i+1][j]-A[i])

print dp[0][N-1]
