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
n=int(input())
X=input()
x=list(map(int,list(X)))
mod=998244353
pow_2=[0]*n
pow_2[n-1]=1
for i in range(1,n):
    pow_2[n-1-i]=(pow_2[n-i]*2)%mod
num=0
for i in range(n):
    if x[i]==1:
        num=(num+pow_2[i])%mod
#約数全列挙
def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    return divisors
li=make_divisors(n)
lst=[]
for i in li:
    if i%2==1 and i>2:
        lst.append(i)
lst.sort(reverse=True)
sum=0
count=0
s=[]
t=[]
for u in lst:
    U=u
    u=n//u
    cnt=0
    for i in range(1,u+1):
        if x[u-i]==1:
            cnt=(cnt+pow_2[-i])%mod
    y=list(X[:u])
    for i in range(u):
        if y[i]=="0":
            y.append("1")
        else:
            y.append("0")
    y="".join(y)
    y*=(U+1)//2
    y=y[:n]
    if y<=X:
        cnt+=1
    for i in range(len(s)):
        if u%s[i]==0:
            cnt-=t[i]
    sum=(sum+(2*cnt*u)%mod)%mod
    s.append(u)
    t.append(cnt)
    count=(count+cnt)%mod
print((((num+1-count)*2*n)%mod+sum)%mod)