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>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,n,m) for(int i=(int)(n); i<=(int)(m); i++)
#define RFOR(i,n,m) for(int i=(int)(n); i>=(int)(m); i--)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

//-------------------------------------------------
//--ModInt
//-------------------------------------------------
const ::std::uint_fast64_t MOD = 998244353;

class mint
{
private:
    using value_type = ::std::uint_fast64_t;
    value_type n;
public:
    mint():n(0){}
    mint(::std::int_fast64_t _n):n(_n<0 ? MOD-(-_n)%MOD : _n%MOD){}
    mint(const mint &m):n(m.n){}

    friend ::std::ostream& operator<<(::std::ostream &os, const mint &a){
		return os << a.n;
	}
    friend ::std::istream& operator>>(::std::istream &is, mint &a){
        value_type temp; is>>temp;
        a = mint(temp);
        return is;
    }

    mint& operator+=(const mint &m){n+=m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator-=(const mint &m){n+=MOD-m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator*=(const mint &m){n=n*m.n%MOD; return *this;}
    mint& operator/=(const mint &m){return *this*=m.inv();}
    mint& operator++(){return *this+=1;}
    mint& operator--(){return *this-=1;}

    mint operator+(const mint &m) const {return mint(*this)+=m;}
    mint operator-(const mint &m) const {return mint(*this)-=m;}
    mint operator*(const mint &m) const {return mint(*this)*=m;}
    mint operator/(const mint &m) const {return mint(*this)/=m;}
    mint operator++(int){mint t(*this); *this+=1; return t;}
    mint operator--(int){mint t(*this); *this-=1; return t;}

    bool operator==(const mint &m) const {return n==m.n;}
    bool operator!=(const mint &m) const {return n!=m.n;}
    mint operator-() const {return mint(MOD-n);}

    mint pow(value_type b) const {
        mint ret(1), m(*this);
        while(b){
            if (b & 1) ret*=m;
            m*=m;
            b>>=1;
        }
        return ret;
    }

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

mint dp[3030][3030];

//-------------------------------------------------

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,S; cin>>N>>S;
    vi a(N);
    rep(i,N) cin>>a[i];
    rep(i,N) dp[i][0]=1;
    rep(i,N)FOR(j,0,S){
        dp[i+1][j] += dp[i][j];
        if (j+a[i]<=S) dp[i+1][j+a[i]] += dp[i][j];
    }
    mint ans=0;
    rep(i,N+1) ans+=dp[i][S];
    cout<<ans<<"\n";
    return 0;
}