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)

// Try Tenka1 Programmer Contest 2019
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    let md = 998244353;
    let n = get!();
    use std::collections::HashMap;
    let mut hm : HashMap<(i64,i64,i64),i64> = HashMap::new();
    let k = (0, 0,0);
    hm.entry(k).or_insert(1);
    for _ in 0.. n {
    	let mut tmp = HashMap::new();
    	let a = get!();
        for (k, v) in hm.iter() {
        	{
        	let k1 = (a+k.0, k.1, k.2);
            let p = tmp.entry(k1).or_insert(0);
            let c = *p;
            *p = (c+v) % md;
            }
            {
        	let k1 = (k.0, a+k.1, 1+k.2);
            let p = tmp.entry(k1).or_insert(0);
            let c = *p;
            *p = (c+v) % md;
            }
        }
        hm = tmp;
    }
    let mut ans = 0;
    for (k, v) in hm.iter() {
    	let r = n - k.2;
    	if k.0 > k.1 && r > 1 {
        	let p = r * (r - 1) / 2 % md;
            let t = v * r % md;
            ans = (ans + t) % md;
        } else if k.0 < k.1 && k.2 > 1 {
        	let p = k.2 * (k.2 - 1) / 2 % md;
            let t = v * r % md;
            ans = (ans + t) % md;
        }
    }
    println!("{:?}", ans);
}