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 PyPy3 (2.4.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 998244353

N, S = MAP()
A = LIST()

# dp0[i][j] := i番目の要素まで見て、総和がjで、使う区間が未確定の状態の通り数
dp0 = list2d(N+1, S+1, 0)
# dp1[i][j] := i番目の要素まで見て、総和がjで、使う区間がL確定済の状態の通り数
dp1 = list2d(N+1, S+1, 0)
# dp2[i][j] := i番目の要素まで見て、総和がjで、使う区間がLR確定済の状態の通り数
dp2 = list2d(N+1, S+1, 0)
dp0[0][0] = 1
for i, a in enumerate(A):
    for j in range(S+1):
        # aを総和に足さない遷移は普通
        dp0[i+1][j] += dp0[i][j]
        dp1[i+1][j] += dp1[i][j]
        dp2[i+1][j] += dp2[i][j]
        # aを総和に足す遷移は、遷移元が状態2と遷移先が状態0は無理
        if j+a <= S:
            # 左端を決める時はそこより手前が自由に使えるのでその分を掛ける
            dp1[i+1][j+a] += dp0[i][j] * (i+1)
            # 右端を決める時はそこより先が自由に使えるのでその分を掛ける
            dp2[i+1][j+a] += dp1[i][j] * (N-i)
            dp1[i+1][j+a] += dp1[i][j] 
            dp2[i+1][j+a] += dp0[i][j] * (i+1) * (N-i)
        dp0[i+1][j] %= MOD
        dp1[i+1][j] %= MOD
        dp2[i+1][j] %= MOD
ans = dp2[N][S]
print(ans)