Effects of Anti Pimples

给定长为 $n$ 的序列 $a$。 你可以选出整数集合 $\{x \mid 1 \le x \le n\}$ 的一个非空子集 $S$，令整数集合 $T = \{k \cdot x \mid x \in S, k \ge 1, k \cdot x \le n\}$。 定义一个选出 $S$ 的方案的分数为 $\max_{i \in T} a_i$。 求所有选出 $S$ 的方案的分数之和。 $1 \le n \le {10^5}$，$0 \le a_i \le {10^5}$。

Chaneka has an array $ [a_1,a_2,\ldots,a_n] $ . Initially, all elements are white. Chaneka will choose one or more different indices and colour the elements at those chosen indices black. Then, she will choose all white elements whose indices are multiples of the index of at least one black element and colour those elements green. After that, her score is the maximum value of $ a_i $ out of all black and green elements. There are $ 2^n-1 $ ways for Chaneka to choose the black indices. Find the sum of scores for all possible ways Chaneka can choose the black indices. Since the answer can be very big, print the answer modulo $ 998\,244\,353 $ .

The first line contains a single integer $ n $ ( $ 1 \leq n \leq 10^5 $ ) — the size of array $ a $ . The second line contains $ n $ integers $ a_1,a_2,a_3,\ldots,a_n $ ( $ 0\leq a_i\leq10^5 $ ).

An integer representing the sum of scores for all possible ways Chaneka can choose the black indices, modulo $ 998\,244\,353 $ .

4
19 14 19 9

265

1
0

0

15
90000 9000 99000 900 90900 9900 99900 90 90090 9090 99090 990 90990 9990 99990

266012571

In the first example, below are the $ 15 $ possible ways to choose the black indices: - Index $ 1 $ is black. Indices $ 2 $ , $ 3 $ , and $ 4 $ are green. Maximum value among them is $ 19 $ . - Index $ 2 $ is black. Index $ 4 $ is green. Maximum value among them is $ 14 $ . - Index $ 3 $ is black. Maximum value among them is $ 19 $ . - Index $ 4 $ is black. Maximum value among them is $ 9 $ . - Indices $ 1 $ and $ 2 $ are black. Indices $ 3 $ and $ 4 $ are green. Maximum value among them is $ 19 $ . - Indices $ 1 $ and $ 3 $ are black. Indices $ 2 $ and $ 4 $ are green. Maximum value among them is $ 19 $ . - Indices $ 1 $ and $ 4 $ are black. Indices $ 2 $ and $ 3 $ are green. Maximum value among them is $ 19 $ . - Indices $ 2 $ and $ 3 $ are black. Index $ 4 $ is green. Maximum value among them is $ 19 $ . - Indices $ 2 $ and $ 4 $ are black. Maximum value among them is $ 14 $ . - Indices $ 3 $ and $ 4 $ are black. Maximum value among them is $ 19 $ . - Indices $ 1 $ , $ 2 $ , and $ 3 $ are black. Index $ 4 $ is green. Maximum value among them is $ 19 $ . - Indices $ 1 $ , $ 2 $ , and $ 4 $ are black. Index $ 3 $ is green. Maximum value among them is $ 19 $ . - Indices $ 1 $ , $ 3 $ , and $ 4 $ are black. Index $ 2 $ is green. Maximum value among them is $ 19 $ . - Indices $ 2 $ , $ 3 $ , and $ 4 $ are black. Maximum value among them is $ 19 $ . - Indices $ 1 $ , $ 2 $ , $ 3 $ , and $ 4 $ are black. Maximum value among them is $ 19 $ . The total sum is $ 19+14+19+9+19+19+19+19+14+19+19+19+19+19+19 = 265 $ .