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
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=998244353
input=lambda :sys.stdin.readline().rstrip()
def divisors(n):
    S,T=[],[]
    for i in range(1,int(n**.5)+1):
        if n%i==0:
            S.append(i)
            T.append(n//i)
    T.reverse()
    return S+T if S[-1]!=T[0] else S+T[1:]

def resolve():
    N=int(input())
    x=input()
    y=''.join('1' if c=='0' else '0' for c in x) # xのbit反転
    D=divisors(N)
    C={d:0 for d in D} # x以下の数字のうち、周期2dの個数
    for d in D:
        if (N//d)%2==0: continue # N/dがevenなら0個
        else:
            z=(x[:d]+y[:d])*((N//d-1)//2)+x[:d]
            C[d]=int(x[:d],2)+1-(x<z)
            C[d]%=MOD
    # ここでC[d]を約数包除で最小周期2dの個数に更新
    for d in D:
        for d1 in D:
            if (d%d1==0 and d!=d1):
                C[d]-=C[d1]
    ans=0
    for d in D:
        ans+=2*d*C[d]%MOD
    print(ans%MOD)
resolve()