#include <bits/stdc++.h>

#define MAXN 3010
#define ll long long
#define mod 998244353

using namespace std;

ll n, tot, ans, s;

ll a[MAXN], f[MAXN]; 

int main() {
	int t;
	scanf("%lld%lld", &n, &s);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = s; j > a[i]; j--) {
			f[j] = (f[j] + f[j - a[i]]) % mod;
		}
		f[a[i]] = (f[a[i]] + i) % mod;
		ans = (ans + (n - i + 1) * f[s] % mod) % mod;
		f[s] = 0;
	}
	cout << (ll)ans << endl;
	return 0;
}
### Prompt

Your task is to create a cpp solution to the following problem:

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 <bits/stdc++.h>

#define MAXN 3010
#define ll long long
#define mod 998244353

using namespace std;

ll n, tot, ans, s;

ll a[MAXN], f[MAXN]; 

int main() {
	int t;
	scanf("%lld%lld", &n, &s);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = s; j > a[i]; j--) {
			f[j] = (f[j] + f[j - a[i]]) % mod;
		}
		f[a[i]] = (f[a[i]] + i) % mod;
		ans = (ans + (n - i + 1) * f[s] % mod) % mod;
		f[s] = 0;
	}
	cout << (ll)ans << endl;
	return 0;
}
```