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.

Examples

Input

3
111


Output

40


Input

6
110101


Output

616


Input

30
001110011011011101010111011100


Output

549320998
import sys
from collections import Counter
readline = sys.stdin.readline

def divi(n):
    res = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            res.append(i)
            if i != n // i:
                res.append(n//i)
    return res

MOD = 998244353
N = int(readline())
X = list(map(int, readline().strip()))
D = divi(N)
C = Counter()
for d in D:
    if N//d % 2 == 0:
        continue
    res = 0
    for x in X[:d]:
        res = (2*res)%MOD
        res += x
    res += 1
    xx = X[:d] + [not x for x in X[:d]]
    for i in range(N):
        x1 = X[i]
        x2 = xx[i%(2*d)]
        if x1 == x2:
            continue
        if x1 < x2:
            res -= 1
            break
        break
    C[2*d] = res%MOD

CL = list(C.items())
CL.sort()
K = len(CL)
Kc, Vc = list(map(list, zip(*CL)))
for i in range(K):
    for j in range(i+1, K):
        if Kc[j] % Kc[i] == 0:
            Vc[j] = (Vc[j] - Vc[i])%MOD
ans = 0
for i in range(K):
    ans = (ans + Kc[i]*Vc[i])%MOD
print(ans)