#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;

signed main(){
	ll N, S;
	cin >> N >> S;
	vector<ll> A(N);
	for(auto &a : A) cin >> a;
	vector<vector<ll>> dp(N+1, vector<ll>(S+1, 0));

	for(int i=0;i<=N;i++) dp[i][0] = 1;

	for(int i=1;i<=N;i++) {
		for(int j=0;j<=S;j++) {
			dp[i][j] += dp[i-1][j];
			if(j >= A[i-1]) dp[i][j] += dp[i-1][j-A[i-1]];
			dp[i][j] %= MOD;
		}
	}
	ll ans = 0;
	for(int i=0;i<=N;i++) {
		ans += dp[i][S];
		ans %= MOD;
	}
	cout << ans << endl;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
using ll = long long;
const ll MOD = 998244353;

signed main(){
	ll N, S;
	cin >> N >> S;
	vector<ll> A(N);
	for(auto &a : A) cin >> a;
	vector<vector<ll>> dp(N+1, vector<ll>(S+1, 0));

	for(int i=0;i<=N;i++) dp[i][0] = 1;

	for(int i=1;i<=N;i++) {
		for(int j=0;j<=S;j++) {
			dp[i][j] += dp[i-1][j];
			if(j >= A[i-1]) dp[i][j] += dp[i-1][j-A[i-1]];
			dp[i][j] %= MOD;
		}
	}
	ll ans = 0;
	for(int i=0;i<=N;i++) {
		ans += dp[i][S];
		ans %= MOD;
	}
	cout << ans << endl;
}
```