Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

4
10 80 90 30

Sample Output 1

10

The game proceeds as follows when the two players play optimally (the element being removed is written bold):

Taro: (10, 80, 90, 30) → (10, 80, 90)

Jiro: (10, 80, 90) → (10, 80)

Taro: (10, 80) → (10)

Jiro: (10) → ()

Here, X = 30 + 80 = 110 and Y = 90 + 10 = 100.

Sample Input 2

3
10 100 10

Sample Output 2

-80

The game proceeds, for example, as follows when the two players play optimally:

Taro: (10, 100, 10) → (100, 10)

Jiro: (100, 10) → (10)

Taro: (10) → ()

Here, X = 10 + 10 = 20 and Y = 100.

Sample Input 3

1
10

Sample Output 3

10

Sample Input 4

10
1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000 1

Sample Output 4

4999999995

The answer may not fit into a 32-bit integer type.

Sample Input 5

6
4 2 9 7 1 5

Sample Output 5

2

The game proceeds, for example, as follows when the two players play optimally:

Taro: (4, 2, 9, 7, 1, 5) → (4, 2, 9, 7, 1)

Jiro: (4, 2, 9, 7, 1) → (2, 9, 7, 1)

Taro: (2, 9, 7, 1) → (2, 9, 7)

Jiro: (2, 9, 7) → (2, 9)

Taro: (2, 9) → (2)

Jiro: (2) → ()

Here, X = 5 + 1 + 9 = 15 and Y = 4 + 7 + 2 = 13.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include<vector>
#include<iostream>
#include<map>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<deque>
#define ll long long
using namespace std;
#define BUF_SIZE 1048576
char _buf[BUF_SIZE],*_is=_buf,*_it=_buf;
inline char rdc(){
	if (_is==_it) _it=(_is=_buf)+fread(_buf,1,BUF_SIZE,stdin);
	return *_is++;
}
inline int rdi(){
	int x=0,f=0;
	register char ch=rdc();
	while ((ch<'0' or ch>'9') and (ch^'-')) ch=rdc();
	if (ch=='-') f=1,ch=rdc();
	while (ch>='0' and ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=rdc();
	return f?-x:x;
}
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
inline void _write(int x){
	write(x);
	putchar(' ');
}
inline void print(int x){
	write(x);
	putchar('\n');
}
const int N=3005;
int dp[N][N];
bool use[N][N];
int a[N];
inline int solve(int l,int r){
	if(l>r) return 0;
	if(use[l][r]) return dp[l][r];
	use[l][r]=1;
	int res=-2147483600;
	res=max(res,a[l]-solve(l+1,r));
	res=max(res,a[r]-solve(l,r-1));
	dp[l][r]=res;
	return res;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n,i;
	cin>>n;
	for(i=1;i<=n;++i) cin>>a[i];
	solve(1,n);
	cout<<dp[1][n]<<endl;
	return 0;
}