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: Time Limit Exceeded)

N = gets.chomp.to_i(10)
X = gets.chomp.to_i(2)

num = 0
ttl = 0
while num <= X do
    moto = num
    taihi = num
	loop do
    #繰り返す処理
    #現在の整数が奇数なら、1を引いて 2で割る。そうでなければ、2 で割って 2N−1を足す。
    	if taihi.modulo(2) == 0 then
                #偶数文字
                 taihi = (taihi / 2) + (2 ** (N - 1))
                 ttl+=1
                 if taihi == moto then
                 break
                 end
        else 
              	#奇数文字
               taihi = (taihi - 1 ) / 2
             	ttl+=1
              if taihi == moto then
              	break 
              end
        end
	end
  num = num + 1
end
puts ttl % 998244353