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: Accepted)

#include <memory.h>

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

#define mod 998244353

long long int result[3010][3010] = {};

int main() {
	int n, s;
	cin >> n >> s;
	long long int ans = 0;
	int a[3000];
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	result[0][a[0]] = 1;
	if (a[0] == s) ans += n;
	for (int i = 1; i < n; i++) {
		result[i][a[i]] += i + 1;
		result[i][a[i]] %= mod;
		if (a[i] == s) {
			ans += (i + 1) * (n - i);
			ans %= mod;
		}
		for (int j = 0; j < s; j++) {
			result[i][j] += result[i - 1][j];
			if (j - a[i] >= 0) result[i][j] += result[i - 1][j - a[i]];
			result[i][j] %= mod;
		}
		if (s - a[i] >= 0) {
			ans += (n - i) * result[i - 1][s - a[i]];
			ans %= mod;
		}
	}
	// for (int i = 0; i < n; i++) {
	// 	for (int j = 0; j <= s; j++) {
	// 		cout << result[i][j] << " ";
	// 	}
	// 	cout << endl;
	// }
	cout << ans << endl;
}