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: Runtime Error)

#include <bits/stdc++.h>
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
constexpr int MOD = 998244353;
constexpr int INF = numeric_limits<int>::max() / 2;
typedef pair<int,int> P;
using Graph = vector<vector<int>>;


signed main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N, S;
	cin >> N >> S;
	int A[N];
	rep(i,N) cin >> A[i];
	int dp[N+1][6500][3]{};
	dp[0][0][0] = 1;
	rep(i,N){
		rep(j,S+1){
			// from 0
			dp[i+1][0][0] = 1;
			dp[i+1][j+A[i]][1] += dp[i][j][0] * (i+1);
			dp[i+1][j+A[i]][2] += dp[i][j][0] * (i+1) * (N-i);

			// from 1
			dp[i+1][j][1] += dp[i][j][1];
			dp[i+1][j+A[i]][1] += dp[i][j][1];
			dp[i+1][j+A[i]][2] += dp[i][j][1] * (N-i);

			// from 2
			dp[i+1][j][2] += dp[i][j][2];

			dp[i+1][j][1] %= MOD;
			dp[i+1][j+A[i]][1] %= MOD;
			dp[i+1][j][2] %= MOD;
			dp[i+1][j+A[i]][2] %= MOD;
		}
	}
	cout << dp[N][S][2] << endl;
 }