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: Runtime Error)

#include<iostream>
#include<vector>
#include<map>
using namespace std;
typedef long long li;
#define repa(i,a,n) for(int i=(a);i<(n);i++)
#define MOD 998244353
#define chmin(a,b) (a>(b)?a=(b),1:0)
#define rep(i,n) for(int i=0;i<(n);i++)
#define df 0
void print(vector<li> v){
  for(int x:v)printf("%d ",x); printf("\n");
} 
template<class T> void print(const T& t){ cout << t << "\n"; }
template<class T, class... Ts> void print(const T& t, const Ts&... ts) { cout << t; if (sizeof...(ts)) cout << " "; print(ts...); }


int main(){
  int n,s; cin >>n >>s;
  vector<int> a(n); rep(i,n) cin >> a[i];
  li cnt =0;
  rep(i,n) if(a[i]==s) (cnt+=(i+1)*(n-i))%=MOD;
  rep(i,n-1){
    if(df)printf("i:%d\n",i);
    vector<li> dp(s+1,0);
    dp[a[i]]=1;
    repa(j,i+1,n){
      if(df)print(dp);
      int b=a[j];
      if(b>=s)continue;
      (cnt+=dp[s-b]*(n-j)%MOD*(i+1)%MOD)%=MOD;
      for(int k=s-b-1;k>=0;k--){
        (dp[k+b]+=dp[k])%=MOD;
      }
    }
  }
  print(cnt);
}