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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

template <int MOD>
struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
    if (val < 0) val += MOD;
  }
  constexpr int getmod() { return MOD; }
  constexpr Fp operator-() const noexcept { return val ? MOD - val : 0; }
  constexpr Fp operator+(const Fp& r) const noexcept { return Fp(*this) += r; }
  constexpr Fp operator-(const Fp& r) const noexcept { return Fp(*this) -= r; }
  constexpr Fp operator*(const Fp& r) const noexcept { return Fp(*this) *= r; }
  constexpr Fp operator/(const Fp& r) const noexcept { return Fp(*this) /= r; }
  constexpr Fp& operator+=(const Fp& r) noexcept {
    val += r.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  constexpr Fp& operator-=(const Fp& r) noexcept {
    val -= r.val;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr Fp& operator*=(const Fp& r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp& operator/=(const Fp& r) noexcept {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp& r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp& r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream& operator<<(ostream& os, const Fp<MOD>& x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD>& a, long long n) noexcept {
    if (n == 0) return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1) t = t * a;
    return t;
  }
};

// 二項係数ライブラリ
template <class T>
struct BiCoef {
  vector<T> fact_, inv_, finv_;
  constexpr BiCoef() {}
  constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
    init(n);
  }
  constexpr void init(int n) noexcept {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    int MOD = fact_[0].getmod();
    for (int i = 2; i < n; i++) {
      fact_[i] = fact_[i - 1] * i;
      inv_[i] = -inv_[MOD % i] * (MOD / i);
      finv_[i] = finv_[i - 1] * inv_[i];
    }
  }
  constexpr T com(int n, int k) const noexcept {
    if (n < k || n < 0 || k < 0) return 0;
    return fact_[n] * finv_[k] * finv_[n - k];
  }
  constexpr T fact(int n) const noexcept {
    if (n < 0) return 0;
    return fact_[n];
  }
  constexpr T inv(int n) const noexcept {
    if (n < 0) return 0;
    return inv_[n];
  }
  constexpr T finv(int n) const noexcept {
    if (n < 0) return 0;
    return finv_[n];
  }
};

const int MOD = 998244353;
using mint = Fp<MOD>;
BiCoef<mint> bc;

// 二項係数Library使用時はmain内で初期化を忘れないこと
// int main(){
//  bc.init(2100000);
// }

int main() {
  ll n, s;
  cin >> n >> s;
  vector<ll> a(n);
  REP(i, n) { cin >> a[i]; }
  mint ans = (mint)0;
  vector<vector<mint>> dp(n + 1, vector<mint>(s + 1, 0));
  dp[0][0] = (mint)1;
  REP(i, n) {
    REP(j, s) {
      dp[i + 1][j] += dp[i][j];
      ll nxt = a[i] + j;
      if (nxt <= s) {
        dp[i + 1][nxt] += dp[i][j];
      }
    }
    ans += dp[i + 1][s] * (mint)(n - i);
  }
  // REP(j, s + 1) { cout << dp[n][j] << endl; }
  reverse(ALL(a));
  vector<vector<mint>> dp2(n + 1, vector<mint>(s + 1, 0));
  dp2[0][0] = (mint)1;
  REP(i, n) {
    REP(j, s) {
      dp2[i + 1][j] += dp2[i][j];
      ll nxt = a[i] + j;
      if (nxt <= s) {
        dp2[i + 1][nxt] += dp2[i][j];
      }
    }
    ans += dp2[i + 1][s] * (mint)(n - i);
  }

  cout << ans << endl;
}