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
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline 

n = int(input())
x = input()

# N のすべての正の約数を求める
# O(sqrt(N))
def divisor_list(N): #約数のリスト
    if N == 1: return [1]
    res = []
    for i in range(1,N):
        if i*i >= N: break
        if N%i == 0:
            res.append(i)
            res.append(N//i)
    if i*i == N: res.append(i)
    return sorted(res)


div = divisor_list(n)

div = [i for i in div if i%2]


ans = []

MOD = 998244353
for i in div:
    per = n//i
    if i == 1:
        less = 0
        eq = 1
        for j in x:
            if j=="1":
                less *= 2
                less += eq
                less %= MOD
            else:
                less *= 2
                less %= MOD
        res = less+1
        #print(less,eq)
        ans.append(res)
    else:
        per = n//i
        less = 0
        eq = 1
        for j in x[:per]:
            if j=="1":
                less *= 2
                less += eq
                less %= MOD
            else:
                less *= 2
                less %= MOD
        res = less

        X = x[:per]
        Y = "".join(["1" if i=="0" else "0" for i in X])
        X = X+Y
        X = (X*i)[:n]
        if X <= x: res += 1
                
        #ans -= res*(2*(n-per))%MOD
        ans.append(res)



for j in range(len(ans)-1,-1,-1):
    for k in range(j):
        if div[j]%div[k] == 0:
            ans[k] -= ans[j]

    
res = 0    
for d,a in zip(div,ans):
    res += a*(n//d)*2%MOD

print(res%MOD)    








