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.util.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int s=sc.nextInt();
		long mod=998244353;
		long[][][] dp=new long[3][n+1][s+1];
		//0は開始、１は継続、２は終了
		for(int i=1; i<=n; i++){
			int a=sc.nextInt();
			dp[0][i][0]=1;
			if(a<=s){
				dp[0][i][a]=1;
			}
			for(int t=s; t>=0; t--){
				dp[1][i][t]+=dp[0][i-1][t]+dp[1][i-1][t];
				if(t>=a){
					dp[1][i][t]+=dp[0][i-1][t-a]+dp[1][i-1][t-a];
				}
				dp[1][i][t]%=mod;
			}
			dp[2][i][s]=(dp[2][i-1][s]+dp[1][i][s]+((a==s)?1:0))%mod;
		}
		System.out.println(dp[2][n][s]);
	}
}