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;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007,MOD2=998244353;



int main() {
    int N,S;
    cin>>N>>S;
    vector<int> A(N);
    rep(i,N){cin>>A.at(i);}
    vector<vector<ll>> dp(N,vector<ll>(S+1));
    //dp.at(0).at(0)=1;
    if(A.at(0)<=S){dp.at(0).at(A.at(0))=1;}
    rep(i,N-1){
        if(A.at(i+1)<=S){dp.at(i+1).at(A.at(i+1))=i+2;}
        for(int j=1;j<S;j++){
            dp.at(i+1).at(j)+=dp.at(i).at(j);
        }
        for(int j=0;j+A.at(i+1)<S+1;j++){
            dp.at(i+1).at(j+A.at(i+1))+=dp.at(i).at(j);
        }
    }
    ll ans=0;
    rep(i,N){
        ll x=dp.at(i).at(S);
        x*=(N-i);
        x%=MOD2;
        ans+=x;
        ans%=MOD2;
    }
    cout<<ans<<endl;
}