Problem:
Score : 800 points

Problem Statement

Given are integers N and X. For each integer k between 0 and X (inclusive), find the answer to the following question, then compute the sum of all those answers, modulo 998244353.

Let us repeat the following operation on the integer k. Operation: if the integer is currently odd, subtract 1 from it and divide it by 2; otherwise, divide it by 2 and add 2^{N-1} to it. How many operations need to be performed until k returns to its original value? (The answer is considered to be 0 if k never returns to its original value.)

Constraints

1 \leq N \leq 2\times 10^5

0 \leq X < 2^N

X is given in binary and has exactly N digits. (In case X has less than N digits, it is given with leading zeroes.)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print the sum of the answers to the questions for the integers between 0 and X (inclusive), modulo 998244353.

Sample Input 1

3
111

Sample Output 1

40

For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

Sample Input 2

6
110101

Sample Output 2

616

Sample Input 3

30
001110011011011101010111011100

Sample Output 3

549320998

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

# 3
# 111 011 001 000 100 110 -> 36
# 010 101 -> 2

N = gets.to_i
X = gets.to_i(2)

upper_bit = 2**(N-1)

ret = Array.new(X+1)

class Node
    attr_accessor :num
end

while !ret.select{|e| e==nil }.empty? do
    target_index = ret.index(nil)
    node = Node.new
    ret[target_index] = node
    ctn = 1
    next_index = target_index.odd? ? target_index/2 : target_index/2 + upper_bit
    while target_index != next_index do
        ret[next_index] = node if next_index <= X
        ctn+=1
        next_index = next_index.odd? ? next_index/2 : next_index/2 + upper_bit
    end
    node.num = ctn
end

puts ret.map{|e| e.num}.inject(:+)