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
def divisors(M):#Mの約数列 O(n^(0.5+e))
    import math
    d=[]
    i=1
    while math.sqrt(M)>=i:
        if M%i==0:
            d.append(i)
            if i**2!=M:
                d.append(M//i)
        i=i+1
    d.sort()
    return d

def Ind(b,n):
    count=0
    while n%b==0:
        count+=1
        n//=b
    return count

N=int(input())
X=input()
x=0
for i in range(N):
    x=2*x+int(X[i])

a=Ind(2,N)
n=N//(2**a)
d=divisors(n)
data={i*2**(a+1):0 for i in d}
mod=998244353

for i in d:
    Q=((2**(i*2**a)+1)*(x+1))//(2**N+1)
    Q%=mod
    data[i*2**(a+1)]+=Q
    data[i*2**(a+1)]%=mod


for i in d:
    for j in d:
        if i>j and i%j==0:
            data[i*2**(a+1)]-=data[j*2**(a+1)]
            data[i*2**(a+1)]%=mod


ans=0
for i in data:
    ans+=data[i]*i
    ans%=mod
print(ans)