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: Accepted)

#include <bits/stdc++.h>
#include <unistd.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define all(v) (v).begin(),(v).end()

#define PII pair<int,int>
#define mp make_pair
#define st first
#define nd second
#define pb push_back
#define lint long long int
#define VI vector<int>

#define debug(x) {cout <<#x <<" = " <<x <<endl; }
#define debug2(x,y) {cerr <<#x <<" = " <<x << ", "<<#y<<" = "<< y <<endl; } 
#define debug3(x,y,z) {cerr <<#x <<" = " <<x << ", "<<#y<<" = "<< y << ", " << #z << " = " << z <<endl; } 
#define debugv(x) {{cout <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<endl; }}
#define debugt(t,n) {{cerr <<#t <<" = "; FOR(it,0,(n)) cerr <<t[it] <<", "; cerr <<endl; }}

#define make( x) int (x); scanf("%d",&(x));
#define make2( x, y) int (x), (y); scanf("%d%d",&(x),&(y));
#define make3(x, y, z) int (x), (y), (z); scanf("%d%d%d",&(x),&(y),&(z));
#define make4(x, y, z, t) int (x), (y), (z), (t); scanf("%d%d%d%d",&(x),&(y),&(z),&(t));
#define makev(v,n) VI (v); FOR(i,0,(n)) { make(a); (v).pb(a);} 
#define IOS ios_base::sync_with_stdio(0)
#define HEAP priority_queue

#define read( x) scanf("%d",&(x));
#define read2( x, y) scanf("%d%d",&(x),&(y));
#define read3(x, y, z) scanf("%d%d%d",&(x),&(y),&(z));
#define read4(x, y, z, t) scanf("%d%d%d%d",&(x),&(y),&(z),&(t));
#define readv(v,n) FOR(i,0,(n)) { make(a); (v).pb(a);}
#define jeb() fflush(stdout);

using namespace std;

const int max_n = 1e5 + 5;

lint dp[3005][3005];

int main () {
	make(n);
	makev(v, n);
	FOR(i,0,n) dp[i][i] = v[i];
	FOR(d,1,n) {
		FOR(i,0,n) {
			if (i+d>n) break;
			dp[i][i+d] = max(v[i]-dp[i+1][i+d], v[i+d]-dp[i][i+d-1]);
		}
	}
	printf("%lld\n", dp[0][n-1]);
}