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 Python (3.4.3) (judge verdict: Wrong Answer)

import numpy as np

N, S = map(int, input().split())
x = [int(x) for x in input().split()]
 
# 状態リストを作成
m = np.zeros((S+1,3))
p = np.zeros((S+1,3))
p[0,0] = 1

# 前状態から次状態の場合数を計算
for i in range(1,N+1):
    
    # x[i] を加えない場合
    m[:,0] = p[:,0]
    m[:,1] = p[:,0] + p[:,1]  # 今回から開始して加えない、もしくは既に開始しているが、加えない
    m[:,2] = p[:,0] + p[:,1] + p[:,2]  # 今回、開始・終了して加えない、今回開始して加えず終了。

    # x[i] を加える場合
    m[x[i-1]:,1] += p[:-x[i-1],0] + p[:-x[i-1],1]
    m[x[i-1]:,2] += p[:-x[i-1],0] + p[:-x[i-1],1]

    p[:,0] = m[:,0] % 998244353
    p[:,1] = m[:,1] % 998244353
    p[:,2] = m[:,2] % 998244353
            
print(int(m[S,2]))