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 "bits/stdc++.h"
using namespace std;

#define fi first
#define se second
#define pb push_back
#define int long long
#define all(x) (x).begin(), (x).end()

using LL = long long;
using LD = long double;
using pii = pair<int, int>;
using vii = vector<pii>;

const int INF = 1e18;
const int MOD = 998244353;
const int N = 3000 + 3;

int n, s;
int a[N];
int dp[N][N];

int solve(int pos, int s) {
    if (s == 0)
        return n - (pos - 1) + 1;

    if (pos > n) 
        return 0;

    if (dp[pos][s] != -1) 
        return dp[pos][s];

    int ans = solve(pos + 1, s);
    if (a[pos] <= s)
        ans += solve(pos + 1, s - a[pos]);

    return dp[pos][s] = ans % MOD;
}

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    
    cin >> n >> s;

    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= s; j++) {
            dp[i][j] = -1;
        }
    }

    int ans = 0;
    for (int i = 1; i <= n; i++) {
        ans += solve(i, s);
        ans %= MOD;
    }

    cout << ans;
}