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)

#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
#define REP(i, n) for(ll i = 0; i < n; ++i)
#define RANGE(i, x, n) for(ll i = x; i < n; ++i)
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;

void accelerate()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
}

const ll mod = 998244353;
const int MAXN = 310;
ll dp[MAXN][MAXN*MAXN];

ll dp2[MAXN][MAXN*MAXN];

ll quickpow(ll x, ll n)
{
    ll ans = 1;
    while(n) {
        if(n&1) {
            ans *= x;
            ans %= mod;
        }
        x *= x;
        x %= mod;
        n >>= 1;
    }
    return ans;
}

int main()
{
    accelerate();
    int n;
    cin >> n;
    vector<int> A(n);
    REP(i, n) {
        cin >> A[i];
    }
    sort(A.begin(), A.end());
    vector<int> sum(n+1);
    REP(i, n) {
        sum[i+1] += sum[i] + A[i];
    }
    REP(i, n+1) {
        if(i == 0) {
            dp[i+1][A[i]] = 1;
            dp[i+1][0] = 2;
        }else {
            REP(j, MAXN*n) {
                dp[i+1][j] += dp[i][j]*2%mod;
                dp[i+1][j] %= mod;
                if(j >= A[i]) dp[i+1][j] += dp[i][j-A[i]];
                dp[i+1][j] %= mod;
            }
        }
    }

    REP(i, n+1) {
        if(i == 0) {
            dp2[i+1][A[i]] = 1;
            dp2[i+1][0] = 1;
        }else {
            REP(j, MAXN*n) {
                dp2[i+1][j] += dp2[i][j]%mod;
                dp2[i+1][j] %= mod;
                if(j >= A[i]) dp2[i+1][j] += dp2[i][j-A[i]];
                dp2[i+1][j] %= mod;
            }
        }
    }


    ll top = sum[n];
    ll ans = 0;
    ll ans2 = 0;
    REP(j, MAXN*n) {
        if(2*j >= top) {
            ans += dp[n][j];
            ans %= mod;
        }
        if(2*j == top){
            cout << dp2[n][j] << endl;
            ans2 += dp2[n][j];
            ans2 %= mod;
        }
    }
    ans = ans*3%mod;
    ans %= mod;
    ans = quickpow(3, n) - ans + ans2*3;
    ans %= mod;
    if(ans < 0) ans += mod;
    cout << ans << endl;
}