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
def get_divisor(n):
    ret = []
    for i in range(3, n + 1, 2):
        if n % i == 0:
            ret.append(i)
    return ret


def get_dec(x, MOD):
    b = 1
    t = 0
    for c in x[::-1]:
        if c == '1':
            t = (t + b) % MOD
        b = b * 2 % MOD
    return t


def count_head_smaller(x, y, d, t, MOD):
    p = x[:t]
    q = y[:t]
    k = get_dec(p, MOD)
    r = (p + q) * (d // 2) + p
    if x >= r:
        k += 1
    return k


def solve(n, x):
    MOD = 998244353
    divisors = get_divisor(n)
    divisors.reverse()
    checked_divisors = {}

    y = ''.join('1' if c == '0' else '0' for c in x)

    ans = 0
    short = 0
    for d in divisors:
        t = n // d
        k = count_head_smaller(x, y, d, t, MOD)

        for pd, pk in checked_divisors.items():
            if pd % d != 0:
                continue
            k -= pk

        ans = (ans + 2 * t * k) % MOD
        short += k
        checked_divisors[d] = k

    t = (get_dec(x, MOD) + 1 - short) % MOD
    ans = (ans + t * 2 * n) % MOD
    return ans


n = int(input())
x = input()
print(solve(n, x))
