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())
a = input()
inva = ''.join(map(lambda x: str(int(x) ^ 1), a))

mod = 998244353
ans = 0
p = 1
num = [0] * -~n
for i in range(n):
    if i:
        num[i] += (num[i - 1] * 2) % mod
    if a[i] == '1':
        num[i] += 1


def get_div(x):
    y = []
    for i in range(1, n):
        if x % i == 0 and (x // i) % 2:
            y.append(i)
    return y


div = get_div(n)
d = [0] * -~n
tot = 0
for i in div:
    now = n // i
    tt = num[i - 1]
    for j in get_div(i):
        tt -= d[j]
    k = (a[:i] + inva[:i]) * (n // (2 * i)) + a[:i]
    d[i] = tt + (a >= k)
    tot += d[i]
d[n] = int(a, 2) + 1 - tot

for i in range(n + 1):
    ans = (ans + d[i] * i * 2) % mod

print(ans % mod)
