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
def fun(i,j):
	# print(i,j)
	if i>j or i>=n or j<0:
		return 0
	elif i==j:
		dp[i][j]=arr[i]
		return dp[i][j]
	else:
		if dp[i][j]==0:
			dp[i][j]=max(arr[i]+min(fun(i+1,j-1),fun(i+2,j)),arr[j]+min(fun(i+1,j-1),fun(i,j-2)))
		return dp[i][j]
n=int(input())
dp=[0]*n
for i in range(n):
	dp[i]=[0]*n
arr=list(map(int,input().split()))

for i in range(n-1,-1,-1):
	for j in range(i,n,1):
		if i==j:
			dp[i][j]=arr[i]
		else:
			dp[i][j]=max(arr[i]-dp[i+1][j],arr[j]-dp[i][j-1])
		# print(dp[i])
print(dp[0][n-1])
# print(2*fun(0,n-1)-sum(arr))