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 OCaml (4.02.3) (judge verdict: Accepted)

let m = 998244353
let ( +^ ) x y = (x + y) mod m
let ( *^ ) x y = (x * y) mod m

let () = Scanf.scanf "%d %d\n" @@ fun n s ->
  let as_ = Array.init n @@ fun _ -> Scanf.scanf "%d " @@ fun a -> a in
  let result = ref 0 in
  let dp = Array.make 3001 0 in
  for i = 0 to n - 1 do
    for j = s downto as_.(i) + 1 do
      dp.(j) <- dp.(j) +^ dp.(j - as_.(i))
    done;
    dp.(as_.(i)) <- dp.(as_.(i)) +^ (i + 1);
    result := !result +^ dp.(s) *^ (n - i);
    dp.(s) <- 0;
  done;
  Printf.printf "%d\n" @@ !result