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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {

  int n,s;
  scanf("%d %d", &n, &s);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  long mod = 998244353;
  long dp[n][s+1][2];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= s; j++) {
      dp[i][j][0] = 0;
      dp[i][j][1] = 0;
    }
  }
  dp[0][0][0] = 1;
  dp[0][0][1] = 1;
  if (a[0] <= s) {
    dp[0][a[0]][0] = 1;
    dp[0][a[0]][1] = 1;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= s; j++) {
      dp[i][j][0] += dp[i-1][j][0];
      if (j+a[i] <= s) {
        dp[i][j+a[i]][0] += dp[i-1][j][0];
        dp[i][j+a[i]][0] %= mod;
      }
    }
    dp[i][0][0]++;
    if (a[i] <= s) {
      dp[i][a[i]][0]++;
      dp[i][a[i]][0] %= mod;
    }
    for (int j = 0; j <= s; j++) {
      dp[i][j][1] = dp[i-1][j][1]+dp[i][j][0];
      dp[i][j][1] %= mod;
    }
  }
  printf("%ld\n", dp[n-1][s][1]);

  return 0;
}