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: Accepted)

import java.io.*;
import java.util.*;


public class Main{
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        // Start writing your solution here. -------------------------------------
        long MOD = 998244353;
        int n = sc.nextInt(), s = sc.nextInt();
        int a[] = new int[n];
        long dp[][] = new long[n + 1][s + 1];
        long res = 0l;
        for(int i = 0; i < n; i++) a[i] = sc.nextInt();
        for(int i = 0;i <= n; i++) dp[i][0] = i + 1;
        for(int i = 1;i <= n; i++)
        {
            for(int j = 1; j < a[i - 1] && j < s; j ++) dp[i][j] = dp[i - 1][j];
            if(a[i - 1] > s) continue;
            for(int j = a[i - 1]; j < s; j ++) dp[i][j] = (dp[i - 1][j - a[i - 1]] + dp[i - 1][j]) % MOD;
            dp[i][s] = dp[i - 1][s - a[i - 1]];
            res += dp[i][s] * (n - i + 1);
        }
        out.println(res % MOD);
        /*
        int n      = sc.nextInt();        // read input as integer
        long k     = sc.nextLong();       // read input as long
        double d   = sc.nextDouble();     // read input as double
        String str = sc.next();           // read input as String
        String s   = sc.nextLine();       // read whole line as String

        int result = 3*n;
        out.println(result);                    // print via PrintWriter
        */

        // Stop writing your solution here. -------------------------------------
        out.close();
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}