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

div = []
i = 1
while i*i <= n:
	if n%i == 0:
		if i%2 == 1:
			div.append(i)
		if i != n//i and (n//i) % 2 == 1:
			div.append(n//i)
	i += 1
div.sort(reverse=True)

ans = dict()

for d in div:
	l = n//d
	blocks = [int(x[i:i+l], 2) for i in range(0, n, l)]
	res = blocks[0]
	same = True
	tmp = blocks[0]
	for i, y in enumerate(blocks[1:]):
		tmp = ((1<<l)-1) ^ tmp
		if tmp < y:
			break
		elif tmp > y:
			same = False
			break
	ans[d] = (res + same) % MOD
	for dd in div:
		if dd == d:
			break
		elif dd % d == 0:
			ans[d] -= ans[dd]
			ans[d] %= MOD

print(sum([(2*n) // k * v for k, v in ans.items()]) % MOD)