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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int long long

signed main(){
  int n;
  cin>>n;
  vector<int> a(n);
  for(int i=0;i<n;i++) cin>>a[i];

  int sum=0;
  for(int i=0;i<n;i++) sum+=a[i];
  
  const int MOD = 998244353;

  vector<int> dp(sum+1,0);
  dp[0]=1;

  for(int i=0;i<n;i++){
    vector<int> nx(sum+1,0);
    for(int j=0;j<=sum;j++){
      nx[j]+=dp[j]*2%MOD;
      if(j+a[i]<=sum) nx[j+a[i]]+=dp[j];
    }
    for(int j=0;j<=sum;j++)
      dp[j]=nx[j]%MOD;
  }

  vector<int> dp2(sum+1,0);
  dp2[0]=1;

  for(int i=0;i<n;i++){
    vector<int> nx(sum+1,0);
    for(int j=0;j<=sum;j++){
      nx[j]+=dp2[j];
      if(j+a[i]<=sum) nx[j+a[i]]+=dp2[j];
    }
    for(int j=0;j<=sum;j++)
      dp2[j]=nx[j]%MOD;
  }

  int ans=0;
  if(sum%2==0){
    ans-=dp2[sum/2]*3%MOD;
    ans+=MOD;
  }
  
  for(int j=(sum+1)/2;j<=sum;j++){
    ans+=dp[j]*3;
    ans%=MOD;
  }
  ans=MOD-ans;

  int po3=1;
  for(int i=0;i<n;i++) po3=po3*3%MOD;
  ans+=po3;
  ans%=MOD;
  
  cout<<ans<<endl;
}