#ABC270H. [ABC270Ex] add 1

[ABC270Ex] add 1

Score : 600600 points

Problem Statement

You are given a tuple of NN non-negative integers A=(A1,A2,,AN)A=(A_1,A_2,\ldots,A_N) such that A1=0A_1=0 and AN>0A_N>0.

Takahashi has NN counters. Initially, the values of all counters are 00. He will repeat the following operation until, for every 1iN1\leq i\leq N, the value of the ii-th counter is at least AiA_i.

Choose one of the NN counters uniformly at random and set its value to 00. (Each choice is independent of others.) Increase the values of the other counters by 11.

Print the expected value of the number of times Takahashi repeats the operation, modulo 998244353998244353 (see Notes).

Notes

It can be proved that the sought expected value is always finite and rational. Additionally, under the Constraints of this problem, when that value is represented as PQ\frac{P}{Q} using two coprime integers PP and QQ, one can prove that there is a unique integer RR such that R×QP(mod998244353)R \times Q \equiv P\pmod{998244353} and 0R<9982443530 \leq R \lt 998244353. Find this RR.

Constraints

  • 2N2×1052\leq N\leq 2\times 10^5
  • 0=A1A2AN10180=A_1\leq A_2\leq \cdots \leq A_N\leq 10^{18}
  • AN>0A_N>0
  • All values in the input are integers.

Input

The input is given from Standard Input in the following format:

NN

A1A_1 A2A_2 \ldots ANA_N

Output

Print the expected value of the number of times Takahashi repeats the operation, modulo 998244353998244353.

2
0 2
6

Let CiC_i denote the value of the ii-th counter.

Here is one possible progression of the process.

  • Set the value of the 11-st counter to 00, and then increase the value of the other counter by 11. Now, (C1,C2)=(0,1)(C_1,C_2)=(0,1).
  • Set the value of the 22-nd counter to 00, and then increase the value of the other counter by 11. Now, (C1,C2)=(1,0)(C_1,C_2)=(1,0).
  • Set the value of the 11-st counter to 00, and then increase the value of the other counter by 11. Now, (C1,C2)=(0,1)(C_1,C_2)=(0,1).
  • Set the value of the 11-st counter to 00, and then increase the value of the other counter by 11. Now, (C1,C2)=(0,2)(C_1,C_2)=(0,2).

In this case, the operation is performed four times.

The probabilities that the process ends after exactly 1,2,3,4,5,1,2,3,4,5,\ldots operation(s) are $0,\frac{1}{4}, \frac{1}{8}, \frac{1}{8}, \frac{3}{32},\ldots$, respectively, so the sought expected value is $2\times\frac{1}{4}+3\times\frac{1}{8}+4\times\frac{1}{8}+5\times\frac{3}{32}+\dots=6$. Thus, 66 should be printed.

5
0 1 3 10 1000000000000000000
874839568