#include <cstdio>

/*
g(L, R) := \prod_{i=L}^{R} (1 + q^{A_i})
f(L, R) = [q^S] g(L, R)
\sum_{1<=L<=R<=N} f(L, R)
p[i] := \sum_{1<=L<=i} g(L, i)
p[i] = (p[i-1] + 1) * (1 + q^{A_i})
*/

constexpr unsigned mod = 998244353;

int n, s;
unsigned dp[3001];

int main(){
  scanf("%d%d", &n, &s);

  long long int ans = 0;
  for(int i = 0; i < n; i++){
    int a;
    scanf("%d", &a);
    dp[0]++;
    for(int j = s; j >= a; j--){
      dp[j] += dp[j - a];
      dp[j] -= mod * (dp[j] >= mod);
    }
    ans += dp[s];
  }
  printf("%d\n", (int) (ans % mod));
  return 0;
}

### Prompt

Your challenge is to write 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 <cstdio>

/*
g(L, R) := \prod_{i=L}^{R} (1 + q^{A_i})
f(L, R) = [q^S] g(L, R)
\sum_{1<=L<=R<=N} f(L, R)
p[i] := \sum_{1<=L<=i} g(L, i)
p[i] = (p[i-1] + 1) * (1 + q^{A_i})
*/

constexpr unsigned mod = 998244353;

int n, s;
unsigned dp[3001];

int main(){
  scanf("%d%d", &n, &s);

  long long int ans = 0;
  for(int i = 0; i < n; i++){
    int a;
    scanf("%d", &a);
    dp[0]++;
    for(int j = s; j >= a; j--){
      dp[j] += dp[j - a];
      dp[j] -= mod * (dp[j] >= mod);
    }
    ans += dp[s];
  }
  printf("%d\n", (int) (ans % mod));
  return 0;
}

```