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: Time Limit Exceeded)

# 解説AC
N,S = map(int, input().split())
A = [int(i) for i in input().split()]

MOD = 998244353

# dp[j][k]
# i: i番目
# j: 区間の総和
# k: 区間の選択状態
# 0: 区間未選択
# 1: 左端のみ選択済み
# 2: 両端が選択済み
dp = [[0] * 3 for _ in range(6000)]
nxt = [[0] * 3 for _ in range(6000)]

# 初期条件
dp[0][0] = 1

# 配るDP
for i, a in enumerate(A):
    for j in range(S + 1):
        # 選択しない
        nxt[j][0] = (nxt[j][0] + dp[j][0]) % MOD
        # 左端を決定
        nxt[j + a][1] = (nxt[j + a][1] + dp[j][0] * (i + 1) % MOD) % MOD
        # 両端を決定
        nxt[j + a][2] = (nxt[j + a][2] + dp[j][0] * (i + 1) * (N - i) % MOD) % MOD

        # 選択しない
        nxt[j][1] = (nxt[j][1] + dp[j][1]) % MOD
        # 選択する
        nxt[j + a][1] = (nxt[j + a][1] + dp[j][1]) % MOD
        # 右端を決定
        nxt[j + a][2] = (nxt[j + a][2] + dp[j][1] * (N - i) % MOD) % MOD

        # 選択しない
        nxt[j][2] = (nxt[j][2] + dp[j][2]) % MOD

    for j in range(S + 1):
        for k in range(3):
            dp[j][k] = nxt[j][k]
    nxt = [[0] * 3 for _ in range(6000)]

print(dp[S][2])