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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce


sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().rstrip().split()
def S(): return sys.stdin.readline().rstrip()
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 998244353

n = I()
x = int(S(), 2)
D = []
for i in range(1, n + 1):
    if n % i == 0 and n // i % 2:
        D += [i]

D.sort()
cnt = [0] * len(D)
ans = 0
for k in range(len(D)):
    bit = x >> (n - D[k])
    cnt[k] = (cnt[k] + bit) % mod
    m = 0
    for i in range(n // D[k]):
        m <<= D[k]
        if i % 2:
            m |= (2 ** D[k] - 1) ^ bit
        else:
            m |= bit
    if m <= x:
        cnt[k] += 1
    ans += cnt[k] * D[k] * 2
    for kk in range(k + 1, len(D)):
        if D[kk] % D[k] == 0:
            cnt[kk] -= cnt[k]
    ans %= mod

print(ans % mod)