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>
#define mp make_pair
#define f first
#define se second
#define pb push_back
#define ms memset
#define MOD 1000000007
#define sp fixed<<setprecision
#define sz sizeof
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
bool pr[1000007];
ll fpow(ll x,ll y){x=x%MOD;ll res=1;while(y){if(y&1)res=res*x;res%=MOD;y=y>>1;x=x*x;x%=MOD;}return res;}
ll inv(ll a,ll m=MOD){ll c=m;ll y=0,x=1;if(m==1)return 0;while(a>1){ll q=a/m;ll t=m;m=a%m,a=t;t=y;y=x-q*y;x=t;}if(x<0)x+=c;return x;}
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	ll n;
	cin>>n;
	ll a[n];
	ll sum=0;
	for(int i=0;i<n;i++)
	{
	cin>>a[i];
	sum+=a[i];
}
ll dp[n][n];
for(int i=n-1;i>=0;i--)
{
	for(int j=i;j<n;j++)
	{
		if(i==j)
		dp[i][j]=a[i];
		else if(j==i-1)
		dp[i][j]=max(a[i],a[j]);
		else
		{
			ll choice1=a[i]+min(dp[i+2][j],dp[i+1][j-1]);
			ll choice2=a[j]+min(dp[i][j-2],dp[i+1][j-1]);
			dp[i][j]=max(choice1,choice2);
		}
	}
}
cout<<2*dp[0][n-1]-sum;
}