Problem:
Score : 600 points

Problem Statement

You are given N integers. The i-th integer is a_i.
Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:

Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.

Constraints

3 \leq N \leq 300

1 \leq a_i \leq 300(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
:
a_N

Output

Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.

Sample Input 1

4
1
1
1
2

Sample Output 1

18

We can only paint the integers so that the lengths of the sides of the triangle will be 1, 2 and 2, and there are 18 such ways.

Sample Input 2

6
1
3
2
3
5
2

Sample Output 2

150

Sample Input 3

20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4

Sample Output 3

563038556

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using static System.Math;

class Program{
    static int n;
    static int[] a;
    static long mod = 998244353;
    static long sum;

    static void Main(){
        n = int.Parse(Console.ReadLine());
        a = new int[n];
        for (int i = 0;i < n;i++) a[i] = int.Parse(Console.ReadLine());
        sum = a.Sum();

        long[,] dp = new long[n + 1,2 * sum + 1];
        dp[0,sum] = 1;
        for (int i = 0;i < n;i++){
            for (int j = a[i];j <= 2 * sum - a[i];j++){
                dp[i + 1,j - a[i]] += 2 * dp[i,j] % mod;
                dp[i + 1,j - a[i]] %= mod;
                dp[i + 1,j + a[i]] += dp[i,j];
                dp[i + 1,j + a[i]] %= mod;
            }
        }

/* 
        for (int i = 0;i <= n;i++){
            for (int j = 0;j <= 2 * sum;j++)
                Console.Write("{0} ",dp[i,j]);
            Console.WriteLine();
        }*/

        long[,] dp2 = new long[n + 1,2 * sum + 1];
        dp2[0,sum] = 1;
        for (int i = 0;i < n;i++){
            for (int j = a[i];j <= 2 * sum - a[i];j++){
                dp2[i + 1,j - a[i]] += dp2[i,j] % mod;
                dp2[i + 1,j - a[i]] %= mod;
                dp2[i + 1,j + a[i]] += dp2[i,j];
                dp2[i + 1,j + a[i]] %= mod;
            }
        }
/* 
        for (int i = 0;i <= n;i++){
            for (int j = 0;j <= 2 * sum;j++)
                Console.Write("{0} ",dp2[i,j]);
            Console.WriteLine();
        }
*/
        long ng = 0;
        for (int i = 0;i < sum;i++){
            ng += (dp[n,i + sum] - 2 * dp2[n,i + sum]) * 3 % mod;
            ng %= mod;
        }

        long ans = (pow(3,n) - 3 * pow(2,n) + 3) % mod;
        ans -= ng;
        ans = (ans + mod) % mod;
        Console.WriteLine(ans);
    }

    static long pow(int a,int k){
        if (k == 0) return 1;
        long t = pow(a, k/2);
        if (k % 2 == 0) return t * t % mod;
        return t * t % mod * a % mod;
    }
}