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 Python (3.4.3) (judge verdict: Wrong Answer)

if __name__ == '__main__':
    import sys
    input = sys.stdin.readline


    nn = int(input())
    x = input()

    ans = 0
    def make_divisors(n):
        divisors = []
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                if i % 2 != 0:
                    divisors.append(i)
                if i != n // i and (n//i)%2 != 0:
                    divisors.append(n//i)

        divisors.sort(reverse = True)
        return divisors

    fact_n = make_divisors(nn)

    used = {}

    ans = 0

    for i in fact_n:
        num_len = int(nn/i)
        base = x[:num_len]
        can_divide = int(base, 2)+1
        for key, value in used.items():
            if num_len%key == 0:
                can_divide -= value
        if x < (base + format(2**num_len-1-int(base,2), 'b'))*int((i-1)/2)+base:
            can_divide -= 1
        used[num_len] = can_divide
        ans += can_divide*num_len*2
        ans = ans%998244353


    print(ans%998244353)