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: Wrong Answer)

# after listening explanation
n = gets.to_i
x = gets.chomp.chars
mod = 998244353
max_n = 200010
exp = []
f = Array.new(max_n*2, 0)
exp[0] = 1
[*1..n].each {|i| exp[i] = 2*exp[i-1]%mod }
k = 1
while k<=n
  if n%k == 0
    l = n/k
    flag = true
    [*0...n].each {|i|
      break if !flag
      f[l] = (f[l]+(i<l ? exp[l-i-1] : 1))%mod if x[i]=='1' && (i<l || x[i-l]=='1')
      flag = false if i>=l && x[i]==x[i-l]
    }
    f[l] = (f[l]+1)%mod if flag
  end
  k+=2
end
ans = 0
[*1..n].each{|i|
  next unless n%i==0 && n/i%2==1 && f[i]
  ans = (ans+2*f[i]*i)%mod
  [*(i*2)..n].each{|j| f[j] = (f[j]+mod-f[i])%mod }
}
p ans