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 (1.42.0) (judge verdict: Wrong Answer)

const MOD: usize = 998_244_353;

fn main() {
    let (n, s): (usize, usize) = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
        )
    };

    let a: Vec<usize> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let iter = buf.split_whitespace();
        iter.map(|x| x.parse().unwrap()).collect()
    };

    let mut ans = 0;
    let mut dp = vec![0; s + 1];
    for i in 0..n {
        if s < a[i] { continue; }
        for j in (0..=(s - a[i])).rev() {
            dp[j + a[i]] = (dp[j + a[i]] + dp[j]) % MOD;
        }
        dp[a[i]] = (dp[a[i]] + (i + 1)) % MOD;
        ans = (ans + dp[s]) % MOD;
    }
    println!("{}", ans);
}