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: Runtime Error)

#include <iostream>
#include <vector>
#include <string>
#include <cstring>
using namespace std;

int MOD = 998244353;

int main(void){

    int N;
    cin >> N;

    int a[N];
    int sum=0;
    long long all=1;
    for(int i=0;i<N;i++){
        cin >> a[i];
        sum += a[i];
        all = (3*all) % MOD;
    }

    long long dp1[N+1][sum+1], dp2[N+1][sum+1];
    memset(dp1, 0, sizeof(dp1));
    memset(dp2, 0, sizeof(dp2));
    dp1[0][0] = dp2[0][0] = 1;

    for(int i=0;i<N;i++){
        for(int j=0;j<=sum;j++){
            if( a[i] <= j ){
                dp1[i+1][j] = (dp1[i][j-a[i]] + 2*dp1[i][j]) % MOD;
                dp2[i+1][j] = (dp2[i][j-a[i]] + dp2[i][j]) % MOD;
            }
            else{
                dp1[i+1][j] = 2*dp1[i][j] % MOD;
                dp2[i+1][j] = dp2[i][j] % MOD;
            }
        }
    }

    long long p1=0;
    for(int i=(sum+1)/2;i<=sum;i++){
        p1 = (p1 + dp1[N][i]) % MOD;
    }
    p1 = 3*p1 % MOD;

    long long p2=0;
    if(sum%2 == 0){
        p2 = dp2[N][sum/2];
    }
    p2 = 3*p2 % MOD;

    long long p3=0;

    long long ans = (all - p1 + p2 - p3) % MOD;

    while (ans < 0) {
        ans += MOD;
    }
    cout << ans << endl;
}