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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>
#include <string>

#define ll long long
using namespace std;
template <class T>ostream &operator<<(ostream &o, const vector<T> &v){
  o << "{";
  for (int i = 0; i < (int)v.size(); i++)
    o << (i > 0 ? ", " : "") << v[i];
  o << "}";
  return o;
}

int main()
{
  int N, MOD = 998244353;
  cin >> N;
  vector<int> a(N);
  vector<vector<int>> dp(N + 1, vector<int>(N * N+1, 0));
  vector<vector<int>> bicol(N + 1, vector<int>(N * N+1, 0));
  dp[0][0] = 1;
  bicol[0][0] = 1;
  for (int i = 0; i < N; i++){
    cin >> a[i];
  }
  int sum = 0;
  for (int i = 0; i < N; i++){
    for (int j = 0; j < N * N; j++){
      dp[i+1][j] += dp[i][j]*2;
      bicol[i+1][j] += bicol[i][j];
      if (j-a[i]>=0) {
        dp[i+1][j] += dp[i][j-a[i]];
        bicol[i+1][j] += bicol[i][j-a[i]];
      }
      dp[i+1][j] %= MOD;
      bicol[i+1][j] %= MOD;
    }
    sum += a[i];
  }

  int ans = 1;
  for (int i = 0; i < N; i++){
    ans *= 3;
    ans %= MOD;
  }

  for (int i = (sum+1)/2; i <= N*N; i++){
    ans += MOD - dp[N][i]*3;
    ans %= MOD;
  }

  if(sum%2 == 0){
    ans += bicol[N][sum/2] * 3;
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}