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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;
int main()
{
	long N, S;
	cin >> N >> S;
	long waru = 998244353;
	vector<vector<long>> dp(S + 1);
	vector<long> pure(S + 1), filled(S+1);//初期値とそこに紐づくパターン数
	vector<long> A(N);
	for (long& a : A) {
		cin >> a;
	}
	long count = 0;
	for (int i = 0; i < N; i++)
	{
		pure = vector<long>(S + 1);
		for (int k = S + 1 - A[i]-1; k >=0 ; k--)
		{
			if (k + A[i] == S) {
				for (int m = 0; m < dp[k].size(); m++)
				{
					count += (dp[k][m] + 1) * (N - i) % waru;
					//break;
				}
			}
			else if (pure[k] == 0 && filled[k]==1) {
				dp[k + A[i]].insert(dp[k + A[i]].end(), dp[k].begin(), dp[k].end());
				pure[k + A[i]] = 1;
				filled[k+A[i]] = 1;
			}


		}
		if (A[i] < S) {
			dp[A[i]].push_back(i);
			filled[A[i]] = 1;
		}
		if (A[i] == S)
		{
			count += (N - i) * (i + 1) % waru;
		}

	}
	cout << count;
	return 0;
}