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>
#define STDIO
#define MOD 998244353
#define testbit(mask,i) (((mask)>>(i))&1)
#define setbit(mask,i) ((mask)|(1<<i))
#define delbit(mask,i) ((mask)^(1<<i))
#define lowbit(x) ((x)&(-x))
#define INF MOD
using namespace std;
int n,s;
int a[3200];
struct segtree{
	long long num[12200][3200];
	void init(){
		memset(num,0,sizeof(num));
	}
	long long query(int x,int l,int r,int ll,int rr,int sum){
		if(r<ll||l>rr)
			return 0;
		else if(ll<=l&&r<=rr){
			return num[x][sum];
		}
		else{
			long long ret=0,ansl,ansr;
			int m=(l+r)/2;
			for(int i=1;i<sum;i++){
				ansl=query(x*2,l,m,ll,rr,i);
				ansr=query(x*2+1,m+1,r,ll,rr,sum-i);
				ret=(ret+(ansl*ansr)%MOD)%MOD;
			}
			ansl=query(x*2,l,m,ll,rr,sum);
			ret=(ret+ansl)%MOD;
			ansr=query(x*2,m+1,r,ll,rr,sum);
			ret=(ret+ansr)%MOD;
			return ret;
		}
	}
	void build(int x,int l,int r){
		if(l==r){
			if(a[l]<=s){
				num[x][a[l]]=1LL;
			}
			return;
		}
		int m=(l+r)/2;
		build(x*2,l,m);
		build(x*2+1,m+1,r);
		for(int i=1;i<=s;i++){
			for(int j=1;j<i;j++){
				num[x][i]=(num[x][i]+(num[x*2][j]*num[x*2+1][i-j])%MOD)%MOD;//左右混合 
			}
			num[x][i]=(num[x][i]+num[x*2][i])%MOD;//全在左边 
			num[x][i]=(num[x][i]+num[x*2+1][i])%MOD;//全在右边 
		}
	}
}seg;
int main(){
	#ifndef STDIO
		freopen("input.in","r",stdin);
		freopen("output.out","w",stdout);
	#endif
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>s;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	seg.init();
	seg.build(1,1,n);
	long long ret=0LL,cand;
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++){
			cand=seg.query(1,1,n,i,j,s);
			ret=(ret+cand)%MOD;
		}
	}
	cout<<ret<<endl;
	return 0;
}