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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cmath>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> ipair;
bool lessPair(const ipair& l, const ipair& r){return l.second < r.second;}
bool morePair(const ipair& l, const ipair& r){return l.second > r.second;}

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const ll MOD = 998244353;
void add(long long &a, long long b) { a += b; if (a >= MOD) a -= MOD; }

ll dp[305][90010];
ll dp2[305][90010];

ll powll(int base, int N) {
   ll ans = 1;
   for (int i = 0; i < N; i++) {
      ans *= base;
      ans %= MOD;
   }
   return ans;
}

int main() {
   int N;
   cin >> N;
   int a[N];
   int sum = 0, sum2 = 0;
   for (int i = 0; i < N; i++) {
      cin >> a[i];
      sum += a[i];
   }
   sum2 = ceil((double)sum / 2.);
   // cout << sum << " " << sum2 << endl;
   sort(a, a + N);
   memset(dp, 0, sizeof(dp));
   memset(dp2, 0, sizeof(dp2));
   dp[0][0] = 1;
   dp2[0][0] = 1;

   for (int i = 0; i < N; i++) {
      for (int v = 0; v + a[i] <= sum; v++) {
         dp[i + 1][v + a[i]] += dp[i][v];
         dp[i + 1][v] += dp[i][v] * 2;
         dp2[i + 1][v + a[i]] += dp2[i][v];
         dp2[i + 1][v] += dp2[i][v];
      }
   }
   /*
   for (int i = 0; i <= N; i++) {
      for (int x = 0; x <= sum; x++) {
         cout << dp[i][x] << " ";
      }
      cout << endl;
   }
   */
   ll ans = 0;
   ans += powll(3, N);
   // cout << ans << endl;
   ll tmp = 0;
   for (int x = sum2; x <= sum; x++) {
      tmp += dp[N][x] * 3;
      tmp %= MOD;
   }
   // cout << tmp << endl;
   ll tmp2 = dp2[N][sum2] * 3;
   tmp2 %= MOD;
   // cout << tmp2 << endl;
   
   ans -= tmp;
   ans %= MOD;
   if (sum % 2 == 0) {
      ans += tmp2;
      ans %= MOD;
   }
   while (ans < 0) ans += MOD;
   cout << ans << endl;
   return 0;
}