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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;

#define ll long long

const ll mod = 998244353;
int N, S, A[3001];
ll dp[3001][3001][3];

void solve() {
	cin >> N >> S;
	for (int i=0; i<N; ++i)
		cin >> A[i+1];
	dp[0][0][0] = 1;
	for (int i=1; i<=N; ++i) {
		for (int j=0; j<=S; ++j) {
			dp[i][j][0] += dp[i-1][j][0];
			dp[i][j][1] += dp[i-1][j][1];
			if (j == 0)
				dp[i][j][1] += dp[i-1][j][0];
			if (j >= A[i]) {
				dp[i][j][0] += dp[i-1][j-A[i]][0];
				dp[i][j][1] += dp[i-1][j-A[i]][1];
				if (j == A[i])
					++dp[i][j][1];	
			}
			dp[i][j][0] %= mod;
			dp[i][j][1] %= mod;
			dp[i][j][2] = (dp[i][j][1] + dp[i-1][j][2])%mod;
		}
	}
	cout << dp[N][S][2] << endl;
}


int main() {
	ios::sync_with_stdio(false);
	solve();
	return 0;
}