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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# dp 状態遷移

n, s = gets.split.map(&:to_i)
nums = gets.split.map(&:to_i)
# dp[上から i 個目の num まで][合計値][未定義、左端定義済、左端右端定義済]の数
dp = (n + 1).times.map { (s + 1).times.map { Array.new(3, 0) } }
# 初期化
# dp[0 桁目まで][合計値は 0][未定義]の出現回数は 1
dp[0][0][0] = 1

# n 個分ループする(i は遷移元の num)
n.times do |i|
  dp_current = dp[i]
  dp_next = dp[i + 1]
  (s + 1).times do |from_sum|
    dp_current_from_sum = dp_current[from_sum]
    dp_next_from_sum = dp_next[from_sum]
    dp_next_from_sum[0] += dp_current_from_sum[0]
    dp_next_from_sum[1] += dp_current_from_sum[1]
    dp_next_from_sum[2] += dp_current_from_sum[2]
    # num を足しても遷移先が s を超えない場合
    to_sum = from_sum + nums[i]
    if to_sum <= s
      dp_next_to_sum = dp_next[to_sum]
      dp_next_to_sum[1] += dp_current_from_sum[1]
      # 左端として採用
      dp_next_to_sum[1] += dp_current_from_sum[0] * (i + 1)
      # 右端として採用
      dp_next_to_sum[2] += dp_current_from_sum[1] * (n - i)
      # 右端および左端として採用
      dp_next_to_sum[2] += dp_current_from_sum[0] * (i + 1) * (n - i)
    end
  end
end

puts dp[n][s][2]