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
mod=998244353
import collections
n=int(input())
x=input()
ans=(2*n*(int(x,2)+1))%mod

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)

  divisors.sort(reverse=True)
  return divisors
D=make_divisors(n)
CT=collections.defaultdict(int)
for d in D:
  if d%2==0 or d==1:
    continue
  else:
    k=n//d
    #下から0がk個、1がk個、、、、を繰り返す数を求める、ｙとする。
    #その周期は2*k
    y=(2**n-2**k)//(2**k+1)

    #ｙ+（ｙ＋１）*yyも同じ周期をもつ
    #x以下の個数を数える
    ct=(int(x,2)-y)//(y+1)+1
    #重複分を差し引く
    Dk=make_divisors(k)
    for dk in Dk:
      if dk<k:
        ct-=CT[dk]
    CT[k]=ct

    #ansはすべてを2nで数えているのでそこから2(n-k)を引く
    ans-=ct*2*(n-k)


print(ans%mod)