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>
using namespace std;

long long dp[3010][3010];
int main() {
  int n, s;
  cin>>n>>s;
  int a[n];
  for (int i=0; i<n+1; i++) cin>>a[i];
  for (int i=0; i<1; i++) {
    for (int j=0; j<s+1; j++) {
      dp[i][j] = 0;
    }
  }
  long long mod = 998244353;
  for (int i=0; i<n; i++) {
    ++dp[i][0];
    for (int j=0; j<s+1; j++) {
      dp[i+1][j] += dp[i][j];
      if (j - a[i] >= 0) {
        dp[i+1][j] += dp[i][j-a[i]];
        dp[i+1][j] %= mod;
      }
    }
  }
  long long ans = 0;
  for (int i=1; i<n+1; i++) {
    ans += dp[i][s];
  }
  cout<<ans<<endl;
}