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;
using ll = long long int;
constexpr ll M = 1e9 + 7;
constexpr ll M2 = 998244353;
inline ll sum_1(const ll n) { return ((n * (n + 1)) / 2) % M; }
inline ll sum_2(const ll n) { return ((n * (n + 1) * (2 * n + 1)) / 6) % M; }
inline ll sum_3(const ll n) { return (((n * (n + 1)) / 2) % M * ((n * (n + 1)) / 2) % M) % M; }
inline ll sum_1(const ll i, const ll j) { return (sum_1(j) - sum_1(i - 1) + M) % M; }
inline ll sum_2(const ll i, const ll j) { return (sum_2(j) - sum_2(i - 1) + M) % M; }
inline ll sum_3(const ll i, const ll j) { return (sum_3(j) - sum_3(i - 1) + M) % M; }

ll pow_f(ll b, ll e) {
    ll ans = 1;
    b = b % M;
    while (e) {
        if (e&1) {
            ans = (ans * b) % M;
        }
        b = (b * b) % M;
        e >>= 1;
    }
    return ans;
}

ll nCr(ll n, ll r) {
    if (r > n || n < 0 || r < 0) { return 0; }
    if (n == r || r == 0) { return 1; }
    r = min(r, n-r);
    ll ans = 1;
    for (ll i = 1; i <= r; ++i) {
        ans = (ans * (n-i+1))/(i); 
    }
    return ans;
}

ll nCr(tuple<ll,ll> t) {
    auto n = get<0>(t);
    auto k = get<1>(t);
    return nCr(n+k-1, k);
}

template<typename T>
void print(T v) {
    cout << v << " ";
}

template<typename T>
void print(vector<T> v) {
    for (auto &i : v) {
        print(i);
    }
    cout << endl;
}

void solve() {
    ll n, s;
    cin >> n >> s;

    vector<ll> A(n);
    for (auto &i : A) {
        cin >> i;
    }

    vector<vector<ll>> dp(n, vector<ll>(s+1, 0));
    vector<vector<ll>> pre(n, vector<ll>(s+1, 0));

    //Base
    for (int i = 0; i < n; ++i) {
        if (A[i] <= s) {
            dp[i][A[i]] = i+1;
            pre[i][A[i]] = i+1;
        }
    }

    //fill
    for (int i = 1; i < n; ++i) {
        for (ll j = 0; j <= s; ++j) {
            if (j-A[i] <= s && j-A[i] >= 0)
                dp[i][j] = (dp[i][j] + pre[i-1][j-A[i]]) % M2;
            pre[i][j] = (pre[i-1][j] + dp[i][j]) % M2;
        }
    }

    // print(dp);   

    ll ans = 0;
    for (ll i = 0; i < n; ++i) {
        ans = (ans + (dp[i][s]*(n-i)) % M2) % M2;
    }

    cout << ans << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifdef D_LOCAL
        freopen("in.in", "r", stdin);
        freopen("out.out", "w", stdout);
    #endif

    solve();

    return 0;
}