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++ (GCC 9.2.1) (judge verdict: Accepted)

#define DEBUG 0
/**
 * File    : F.cpp
 * Author  : Kazune Takahashi
 * Created : 6/19/2020, 11:32:52 PM
 * Powered by Visual Studio Code
 */
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>
// ----- boost -----
#include <boost/integer/common_factor_rt.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/rational.hpp>
// ----- using directives and manipulations -----
using namespace std;
using boost::rational;
using boost::integer::gcd; // for C++14 or for cpp_int
using boost::integer::lcm; // for C++14 or for cpp_int
using boost::multiprecision::cpp_int;
using ll = long long;
using ld = long double;
template <typename T>
using max_heap = priority_queue<T>;
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
// ----- constexpr for Mint and Combination -----
// constexpr ll MOD{1'000'000'007LL};
constexpr ll MOD{998'244'353LL}; // be careful
constexpr ll MAX_SIZE{3'000'010LL};
// constexpr ll MAX_SIZE{30'000'010LL}; // if 10^7 is needed
// ----- ch_max and ch_min -----
template <typename T>
bool ch_max(T &left, T right)
{
  if (left < right)
  {
    left = right;
    return true;
  }
  return false;
}
template <typename T>
bool ch_min(T &left, T right)
{
  if (left > right)
  {
    left = right;
    return true;
  }
  return false;
}
// ----- Mint -----
template <ll MOD = MOD>
class Mint
{
public:
  ll x;
  Mint() : x{0LL} {}
  Mint(ll x) : x{(x % MOD + MOD) % MOD} {}
  Mint operator-() const { return x ? MOD - x : 0; }
  Mint &operator+=(Mint const &a)
  {
    if ((x += a.x) >= MOD)
    {
      x -= MOD;
    }
    return *this;
  }
  Mint &operator-=(Mint const &a) { return *this += -a; }
  Mint &operator++() { return *this += 1; }
  Mint operator++(int)
  {
    Mint tmp{*this};
    ++*this;
    return tmp;
  }
  Mint &operator--() { return *this -= 1; }
  Mint operator--(int)
  {
    Mint tmp{*this};
    --*this;
    return tmp;
  }
  Mint &operator*=(Mint const &a)
  {
    (x *= a.x) %= MOD;
    return *this;
  }
  Mint &operator/=(Mint const &a)
  {
    Mint b{a};
    return *this *= b.power(MOD - 2);
  }
  Mint operator+(Mint const &a) const { return Mint(*this) += a; }
  Mint operator-(Mint const &a) const { return Mint(*this) -= a; }
  Mint operator*(Mint const &a) const { return Mint(*this) *= a; }
  Mint operator/(Mint const &a) const { return Mint(*this) /= a; }
  bool operator<(Mint const &a) const { return x < a.x; }
  bool operator<=(Mint const &a) const { return x <= a.x; }
  bool operator>(Mint const &a) const { return x > a.x; }
  bool operator>=(Mint const &a) const { return x >= a.x; }
  bool operator==(Mint const &a) const { return x == a.x; }
  bool operator!=(Mint const &a) const { return !(*this == a); }
  Mint power(ll N) const
  {
    if (N == 0)
    {
      return 1;
    }
    else if (N % 2 == 1)
    {
      return *this * power(N - 1);
    }
    else
    {
      Mint half = power(N / 2);
      return half * half;
    }
  }
};
template <ll MOD>
Mint<MOD> operator+(ll lhs, Mint<MOD> const &rhs) { return rhs + lhs; }
template <ll MOD>
Mint<MOD> operator-(ll lhs, Mint<MOD> const &rhs) { return -rhs + lhs; }
template <ll MOD>
Mint<MOD> operator*(ll lhs, Mint<MOD> const &rhs) { return rhs * lhs; }
template <ll MOD>
Mint<MOD> operator/(ll lhs, Mint<MOD> const &rhs) { return Mint<MOD>{lhs} / rhs; }
template <ll MOD>
istream &operator>>(istream &stream, Mint<MOD> &a) { return stream >> a.x; }
template <ll MOD>
ostream &operator<<(ostream &stream, Mint<MOD> const &a) { return stream << a.x; }
// ----- Combination -----
template <ll MOD = MOD, ll MAX_SIZE = MAX_SIZE>
class Combination
{
public:
  vector<Mint<MOD>> inv, fact, factinv;
  Combination() : inv(MAX_SIZE), fact(MAX_SIZE), factinv(MAX_SIZE)
  {
    inv[1] = 1;
    for (auto i{2LL}; i < MAX_SIZE; i++)
    {
      inv[i] = (-inv[MOD % i]) * (MOD / i);
    }
    fact[0] = factinv[0] = 1;
    for (auto i{1LL}; i < MAX_SIZE; i++)
    {
      fact[i] = Mint<MOD>(i) * fact[i - 1];
      factinv[i] = inv[i] * factinv[i - 1];
    }
  }
  Mint<MOD> operator()(int n, int k)
  {
    if (n >= 0 && k >= 0 && n - k >= 0)
    {
      return fact[n] * factinv[k] * factinv[n - k];
    }
    return 0;
  }
  Mint<MOD> catalan(int x, int y)
  {
    return (*this)(x + y, y) - (*this)(x + y, y - 1);
  }
};
// ----- for C++14 -----
using mint = Mint<MOD>;
using combination = Combination<MOD, MAX_SIZE>;
// ----- for C++17 -----
template <typename T, typename enable_if<is_integral<T>::value>::type * = nullptr>
size_t popcount(T x) { return bitset<64>(x).count(); }
size_t popcount(string const &S) { return bitset<200010>{S}.count(); }
// ----- Infty -----
template <typename T>
constexpr T Infty() { return numeric_limits<T>::max(); }
template <typename T>
constexpr T mInfty() { return numeric_limits<T>::min(); }
// ----- frequently used constexpr -----
// constexpr double epsilon{1e-10};
// constexpr ll infty{1'000'000'000'000'010LL}; // or
// constexpr int infty{1'000'000'010};
// constexpr int dx[4] = {1, 0, -1, 0};
// constexpr int dy[4] = {0, 1, 0, -1};
// ----- Yes() and No() -----
void Yes()
{
  cout << "Yes" << endl;
  exit(0);
}
void No()
{
  cout << "No" << endl;
  exit(0);
}

// ----- Solve -----

class Solve
{

public:
  Solve()
  {
  }

  void flush()
  {
  }

private:
};

// ----- 2D, 3D, 4D vectors -----
// Referring to ymatsux-san's source code: https://atcoder.jp/contests/abc138/submissions/7018300

template <typename T>
vector<vector<T>> Make2DVector(size_t d0, size_t d1, T v = T{})
{
  return vector<vector<T>>(d0, vector<T>(d1, v));
}

template <typename T>
vector<vector<vector<T>>> Make3DVector(size_t d0, size_t d1, size_t d2, T v = T{})
{
  return vector<vector<vector<T>>>(d0, Make2DVector(d1, d2, v));
}

template <typename T>
vector<vector<vector<vector<T>>>> Make4DVector(size_t d0, size_t d1, size_t d2, size_t d3, T v = T{})
{
  return vector<vector<vector<vector<T>>>>(d0, Make3DVector(d1, d2, d3, v));
}

// ----- main() -----

/*
int main()
{
  Solve solve;
  solve.flush();
}
*/

int main()
{
  int N, S;
  cin >> N >> S;
  vector<int> A(N);
  for (auto i{0}; i < N; ++i)
  {
    cin >> A[i];
  }
  auto dp{Make3DVector<mint>(N + 1, S + 1, 3, 0)};
  dp[0][0][0] = 1;
  for (auto i{0}; i < N; ++i)
  {
    for (auto j{0}; j <= S; ++j)
    {
      for (auto k{0}; k < 3; ++k)
      {
        for (auto nk{0}; nk < 3; ++nk)
        {
          for (auto l{0}; l < 2; ++l)
          {
            int nj{l == 0 ? j : j + A[i]};
            if (nj > S)
            {
              continue;
            }
            if (k > nk)
            {
              continue;
            }
            if (l == 1)
            {
              if ((k == 0 && nk == 0) || (k == 2 && nk == 2))
              {
                continue;
              }
            }
            dp[i + 1][nj][nk] += dp[i][j][k];
          }
        }
      }
    }
  }
  cout << dp[N][S][2] << endl;
}