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 <cstdio>
#include <algorithm>  
#include <bitset>
#include <cassert>
#include <ctime>  
#include <cmath>  
#include <cstdlib>  
#include <cstring>  
#include <functional>  
#include <iostream>  
#include <deque>  
#include <stack>
#include <queue>  
#include <vector>  
#include <map>  
#include <set>  
#include <string>  
#include <unordered_map>  
#include <unordered_set>  
#pragma warning(disable:4996)

using namespace std;

#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define endl '\n'

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, int> pdi;

const int INF = 1e9;
const int MAXN = 3e3 + 5;
const ll MOD = 998244353;

ll N, S, A[MAXN], ans;
ll D[MAXN][MAXN];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> N >> S;
	for(int i=1; i<=N; i++) cin >> A[i];

	for(int i=1; i<=N; i++) {
		D[i-1][0] = i;
		for(int w=1; w<=S; w++) {
			if(w >= A[i]) D[i][w] = (D[i-1][w] + D[i-1][w-A[i]]) % MOD;
			else D[i][w] = D[i-1][w];
		}
		ans = (ans + D[i][S]) % MOD;
	}
	cout << ans;
}