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: Runtime Error)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		int N = int.Parse(Console.ReadLine());
		int[]A = new int[N];
		for(var i=0;i<N;i++){
			A[i] = int.Parse(Console.ReadLine());
		}
		Array.Sort(A);
		var dic = new Dictionary<long,long>();
		var list = new List<long[]>();
		list.Add(new long[3]{A[0],0,0});
		dic[A[0]*45000*45000] = 1;
		list.Add(new long[3]{0,A[0],0});
		dic[A[0]*45000] = 1;
		list.Add(new long[3]{0,0,A[0]});
		dic[A[0]] = 1;
		long mod = 998244353;
		for(var i=1;i<N;i++){
			var list2 = new List<long[]>(list);
			list.Clear();
			for(var j=0;j<list2.Count;j++){
				long a1 = list2[j][0];
				long a2 = list2[j][1];
				long a3 = list2[j][2];
				if(a1>=45000 || a2>=45000 || a3>=45000){
					continue;
				}
				long b = dic[a1*45000*45000+a2*45000+a3];
				long k1 = (a1+A[i])*45000*45000+a2*45000+a3;
				long k2 = a1*45000*45000+(a2+A[i])*45000+a3;
				long k3 = a1*45000*45000+a2*45000+(a3+A[i]);
				if(dic.ContainsKey(k1)){
					dic[k1] = (dic[k1] + b) % mod;
				} else {
					dic[k1] = b;
					list.Add(new long[3]{a1+A[i],a2,a3});
				}
				if(dic.ContainsKey(k2)){
					dic[k2] = (dic[k2] + b) % mod;
				} else {
					dic[k2] = b;
					list.Add(new long[3]{a1,a2+A[i],a3});
				}
				if(dic.ContainsKey(k3)){
					dic[k3] = (dic[k3] + b) % mod;
				} else {
					dic[k3] = b;
					list.Add(new long[3]{a1,a2,a3+A[i]});
				}
			}
		}
		long ans = 0;
		for(var i=0;i<list.Count;i++){
			if(list[i][0] > 0 && list[i][1] > 0 && list[i][2] > 0){
				if(list[i][0]+list[i][1]>list[i][2] && list[i][0]+list[i][2]>list[i][1] && list[i][1]+list[i][2]>list[i][0]){
					ans += dic[list[i][0]*45000*45000+list[i][1]*45000+list[i][2]];
					ans %= mod;
				}
			}
		}
		Console.WriteLine(ans);
	}
}