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
import java.util.*;
import java.io.*;

public class Main {
 	
  
  public static void main(String abc[]) {
     Main ans = new Main();
     System.out.println(ans.solve());
  }
  
  private long solve() {
     Scanner sc = new Scanner(System.in);
     int N = sc.nextInt();
     long [] array = new long [N+1];
     int i=0;
     long sum=0;
     for(i=1;i<=N;i++) {
        array[i] = sc.nextLong();
        sum += array[i];
     }
     //System.out.println(Arrays.toString(array));
    
     long [][][] dp =new long[3001][3001][2];
     for(i=0;i<3001;i++) {
        for(int j=0;j<3001;j++) {
            dp[i][j][0] = -1;
            dp[i][j][1] = -1;
        }
     }
    
     long X =  solve(dp, array, 1, 1, N);
     long Y = sum-X;
     return X-Y;
  }
  
  private long solve(long[][][] dp, long[] array, int turn, int i, int j) {
     if(i>j) return 0;
     if (dp[i][j][turn] !=-1) return dp[i][j][turn];
     if (turn==1) {
         long left = array[i] + solve(dp, array, 0, i+1, j);
         long right = array[j] + solve(dp, array, 0, i, j-1);
         dp[i][j][turn] = Math.max(left,right);
     } else {
         long left = solve(dp, array, 1, i+1, j);
         long right = solve(dp, array, 1, i, j-1);
         dp[i][j][turn] = Math.min(left, right);
     }
     return dp[i][j][turn];
  }
}
