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>
using namespace std;
using ll = long long int;
const ll mod = 998244353;
vector<int> a(4000);
int n,s;
ll dp[4000][4000][3];
int ways(int l,int r,int S){
    if(S == 0){
        return 1;
    }
    if(r == l){
        return 0;
    }
    return ways(l,r-1,S) + (S>=a[r-1]?ways(l,r-1,S-a[r-1]):0);
}



int main(){
    cin >> n >> s;
    for(int i = 0;i<n;++i){
        cin >> a[i];
    }
   
    /*
     int ans  = 0;
    for(int i = 0;i<n;++i){
        for(int j = i;j < n;++j){
            ans += ways(i,j+1,s);
            if(ans >= mod){
                ans-=mod;
            }
        }
    }
    
    cout << ans <<"\n";
    */
   auto add =[&](ll &x,ll y){
       x += y;
       x %= mod;
   };
   
   memset(dp,0LL,sizeof dp);
   dp[0][0][0] = 1;
   for(int i = 0;i<n;++i){
       for(int j = 0;j<=s;++j){
            add(dp[i+1][j][0],dp[i][j][0]);
            add(dp[i+1][j][1],dp[i][j][0]+dp[i][j][1]);
            add(dp[i+1][j][2],dp[i][j][0]+dp[i][j][1]+dp[i][j][2]);
            if(j + a[i] <= s){
                add(dp[i+1][j+a[i]][1],dp[i][j][0] + dp[i][j][1]);
                add(dp[i+1][j+a[i]][2],dp[i][j][0] + dp[i][j][1]);
            }  
       }
   }
   cout << dp[n][s][2] <<"\n";
    return 0;    
}