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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    static int[][] arr;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int s = sc.nextInt();
        int[] A = new int[n];
        int mod = 998244353;
        arr = new int[n + 1][n + 1];
        long ans = 0;
        for(int i = 0; i < n; i++) A[i] = Integer.parseInt(sc.next());
        for(int range = 0; range <= n; range++) {
            for(int i = 1; i + range <= n; i++) {
                int sum = 0;
                for(int j = i; j <= i + range; j++) sum += A[j - 1];
                if(sum == s) {
                    arr[i][i + range]++;
                }
                if(i != 1 && i + range != n) arr[i - 1][i + range + 1] += arr[i][i + range] % mod;
                if(i != 1) arr[i - 1][i + range] += arr[i][i + range] % mod;
                if(i + range != n) arr[i][i + range + 1] += arr[i][i + range] % mod;
            }
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                ans += arr[i][j] % mod;
            }
        }
        System.out.println(ans);
    }
}