#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
#define F(i,x,y) for (int i=x; i<=y; i++)
#define D(i,x,y) for (int i=x; i>=y; i--)
using namespace std;
const int N=3333;
ll f[N];
int a[N];
int n,k;
ll p=998244353;
int main(){
	scanf("%d%d",&n,&k);
	F(i,1,n) scanf("%d",&a[i]);
	ll ans=0;
	F(i,1,n){
		f[0]=i;
		D(j,k,a[i]) f[j]=(f[j]+f[j-a[i]])%p;
		ans=(ans+f[k])%p;
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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.

Examples

Input

3 4
2 2 4


Output

5


Input

5 8
9 9 9 9 9


Output

0


Input

10 10
3 1 4 1 5 9 2 6 5 3


Output

152

### Response

```cpp
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
#define F(i,x,y) for (int i=x; i<=y; i++)
#define D(i,x,y) for (int i=x; i>=y; i--)
using namespace std;
const int N=3333;
ll f[N];
int a[N];
int n,k;
ll p=998244353;
int main(){
	scanf("%d%d",&n,&k);
	F(i,1,n) scanf("%d",&a[i]);
	ll ans=0;
	F(i,1,n){
		f[0]=i;
		D(j,k,a[i]) f[j]=(f[j]+f[j-a[i]])%p;
		ans=(ans+f[k])%p;
	}
	printf("%lld\n",ans);
	return 0;
}
```