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 Rust (1.15.1) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let n = read::<usize>();
    let mut a = vec![0usize; n];
    let big_int = 998244353u64;
    for i in 0..n {
        a[i] = read::<usize>();
    }
    let sum_length = a.iter().sum::<usize>();

    let mut dp = vec![vec![0u64; sum_length + 1]; n + 1];
    dp[0][0] = 1;
    for i in 0..n {
        for j in (0..sum_length + 1).rev() {
            // not select
            dp[i + 1][j] += 2 * dp[i][j];
            dp[i + 1][j] %= big_int;
            // select
            if j + a[i] <= sum_length {
                dp[i + 1][j + a[i]] += dp[i][j];
                dp[i + 1][j + a[i]] %= big_int;
            }
        }
    }
    let mut sum = 0;
    for j in (sum_length + 1) / 2..sum_length + 1 {
        sum += dp[n][j];
        sum %= big_int;
    }
    sum *= 3;
    sum %= big_int;
    let mut ans = 1;
    for i in 0..n {
        ans *= 3;
        ans %= big_int;
    }
    ans += big_int - sum;
    ans %= big_int;
    //println!("{:?}", dp);
    println!("{}", ans);
    //println!("{}", sum);
}