Problem:
Score : 600 points

Problem Statement

Given are a sequence of N integers A_1, A_2, \ldots, A_N and a positive integer S.

For a pair of integers (L, R) such that 1\leq L \leq R \leq N, let us define f(L, R) as follows:

f(L, R) is the number of sequences of integers (x_1, x_2, \ldots , x_k) such that L \leq x_1 < x_2 < \cdots < x_k \leq R and A_{x_1}+A_{x_2}+\cdots +A_{x_k} = S.

Find the sum of f(L, R) over all pairs of integers (L, R) such that 1\leq L \leq R\leq N. Since this sum can be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N \leq 3000

1 \leq S \leq 3000

1 \leq A_i \leq 3000

Input

Input is given from Standard Input in the following format:

N S
A_1 A_2 ... A_N

Output

Print the sum of f(L, R), modulo 998244353.

Sample Input 1

3 4
2 2 4

Sample Output 1

5

The value of f(L, R) for each pair is as follows, for a total of 5.

f(1,1) = 0

f(1,2) = 1 (for the sequence (1, 2))

f(1,3) = 2 (for (1, 2) and (3))

f(2,2) = 0

f(2,3) = 1 (for (3))

f(3,3) = 1 (for (3))

Sample Input 2

5 8
9 9 9 9 9

Sample Output 2

0

Sample Input 3

10 10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

152

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn vec<T, F: FnMut(usize) -> T>(n: usize, mut f: F) -> Vec<T> {
    (0..n).map(|i| f(i)).collect()
}

const MODULO: usize = 998244353;

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let s: usize = scan(tokens);
    let a: Vec<usize> = vec(n, |_| scan(tokens));
    let mut count: Vec<usize> = vec![0; s];
    let mut ans: usize = 0;
    for i in 0..n {
        if a[i] > s {
            continue;
        }
        ans = (ans + count[s - a[i]] * (n - i)) % MODULO;
        for x in (a[i] + 1..s).rev() {
            count[x] = (count[x] + count[x - a[i]]) % MODULO;
        }
        if a[i] == s {
            ans = (ans + (i + 1) * (n - i)) % MODULO;
        } else {
            count[a[i]] = (count[a[i]] + i + 1) % MODULO;
        }
    }
    println!("{}", ans);
}