// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)

#define rep2(i, x, n) for(int i = x, i##_len = (n); i < i##_len; ++i)

#define all(n) begin(n), end(n)

using ll = long long;

using P = pair<int, int>;

using vi = vector<int>;

using vl = vector<ll>;

using vs = vector<string>;

using vc = vector<char>;

using vb = vector<bool>;

using vd = vector<double>;

vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};



const int mod = 998244353;

struct mint {

  ll x;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint& operator+=(const mint a) {

    if((x += a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator-=(const mint a) {

    if((x += mod - a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator*=(const mint a) {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const {

    mint res(*this);

    return res += a;

  }

  mint operator-(const mint a) const {

    mint res(*this);

    return res -= a;

  }

  mint operator*(const mint a) const {

    mint res(*this);

    return res *= a;

  }

  mint pow(ll t) const {

    if(!t) return 1;

    mint a = pow(t >> 1);

    a *= a;

    if(t & 1) a *= *this;

    return a;

  }

  // for prime mod

  mint inv() const { return pow(mod - 2); }

  mint& operator/=(const mint a) { return (*this) *= a.inv(); }

  mint operator/(const mint a) const {

    mint res(*this);

    return res /= a;

  }

};

mint fact(int n) {

  if(n == 1 || n == 0) return 1;

  mint res = n;

  return res * fact(n - 1);

}



int main() {

  ll n, s;

  cin >> n >> s;

  vl a(n);

  rep(i, n) cin >> a[i];

  vector<vector<mint>> dp(n + 1, vector<mint>(s + 3010, 0));

  rep(i, n + 1) dp[i][0] = 1;

  rep(i, n) rep(j, s + 1) {

    dp[i + 1][j] += dp[i][j];

    dp[i + 1][j + a[i]] += dp[i][j];

  }

  mint ans = 0;

  rep(i, n + 1) ans += dp[i][s];

  cout << ans.x << endl;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)

#define rep2(i, x, n) for(int i = x, i##_len = (n); i < i##_len; ++i)

#define all(n) begin(n), end(n)

using ll = long long;

using P = pair<int, int>;

using vi = vector<int>;

using vl = vector<ll>;

using vs = vector<string>;

using vc = vector<char>;

using vb = vector<bool>;

using vd = vector<double>;

vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};



const int mod = 998244353;

struct mint {

  ll x;  // typedef long long ll;

  mint(ll x = 0) : x((x % mod + mod) % mod) {}

  mint operator-() const { return mint(-x); }

  mint& operator+=(const mint a) {

    if((x += a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator-=(const mint a) {

    if((x += mod - a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator*=(const mint a) {

    (x *= a.x) %= mod;

    return *this;

  }

  mint operator+(const mint a) const { return mint(*this) += a; }

  mint operator-(const mint a) const { return mint(*this) -= a; }

  mint operator*(const mint a) const { return mint(*this) *= a; }

  mint pow(ll t) const {

    if(!t) return 1;

    mint a = pow(t >> 1);

    a *= a;

    if(t & 1) a *= *this;

    return a;

  }

  // for prime mod

  mint inv() const { return pow(mod - 2); }

  mint& operator/=(const mint a) { return *this *= a.inv(); }

  mint operator/(const mint a) const { return mint(*this) /= a; }

};

istream& operator>>(istream& is, const mint& a) { return is >> a.x; }

ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

mint fact(int n) {

  if(n == 1 || n == 0) return 1;

  mint res = n;

  return res * fact(n - 1);

}



int main() {

  ll n, s;

  cin >> n >> s;

  vl a(n);

  rep(i, n) cin >> a[i];

  mint dp[s + 1] = {}, ans = 0;

  rep(i, n) {

    dp[0] += 1;

    for(int j = s; j >= 0; j--) {

      int nj = j + a[i];

      if(nj <= s) dp[nj] += dp[j];

    }

    ans += dp[s];

  }

  cout << ans << endl;

}

// Fastest known solution for this problem: