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
#include <bits/stdc++.h>

const int N = 2e5 + 5, djq = 998244353;

int n, tot, p[N], p2[N], f[N], ans;
char s[N];

char inv(char c) {return c == '0' ? '1' : '0';}

char conv(int d, int i) {return i / d & 1 ? inv(s[i % d]) : s[i % d];}

int calc(int d)
{
	int res = 0;
	for (int i = 0; i < d; i++) if (s[i] == '1')
		res = (res + p2[d - i - 1]) % djq;
	for (int i = 0; i < n; i++)
		if (conv(d, i) < s[i]) return (res + 1) % djq;
		else if (conv(d, i) > s[i]) return res;
	return (res + 1) % djq;
}

int main()
{
	scanf("%d%s", &n, s);
	int S = sqrt(n); p2[0] = 1;
	for (int i = 1; i <= n; i++) p2[i] = (p2[i - 1] << 1) % djq;
	for (int i = 1; i <= S; i++)
	{
		if (n % i) continue; p[++tot] = i;
		if (i < n / i) p[++tot] = n / i;
	}
	std::sort(p + 1, p + tot + 1);
	for (int i = 1; i <= tot; i++)
	{
		int x = p[i]; if (!(n / x & 1)) continue;
		f[x] = calc(x);
		for (int j = 1; j < i; j++)
			if (x % p[j] == 0) f[x] = (f[x] - f[p[j]] + djq) % djq;
		ans = (2ll * x * f[x] + ans) % djq;
	}
	return std::cout << ans << std::endl, 0;
}