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 mod = 998244353;
const int MAX_N = int(2e5) + 10;

int n;
char s[MAX_N];
int pre[MAX_N], f[MAX_N];

int check(int x) {
	int cur = 1, flag = 0;
	for (int i = 1; i <= n; ++i) {
		int t1 = (s[cur] - '0') ^ flag, t2 = s[i] - '0';
		if (t1 > t2)
			return 0;
		else {
			if (t1 < t2) return 1;
			++cur;
			if (cur > x) cur = 1, flag ^= 1;
		}
	}
	return 1;
}

int get(int x) {
	if ((n / x) & 1)
		return pre[x] + check(x);
	else
		return 0;
}

int main() {
	scanf("%d%s", &n, s + 1);
	int tot = 0;
	for (int i = 1; i <= n; ++i) {
		tot = 1LL * tot * 2 % mod;
		if (s[i] == '1')
			tot = (tot + 1) % mod;
		pre[i] = tot;
	}

	for (int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			f[i] = get(i);
			if (n / i != i)
				f[n / i] = get(n / i);
		}
	}

	for (int i = 1; i <= n; ++i) {
		for (int j = i + i; j <= n; j += i) {
			f[j] = (f[j] - f[i] + mod) % mod;
		}
	}

	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		if (n % i == 0) {
			ans = (ans + 1LL * f[i] * 2 * i % mod) % mod;
		}
	}
	printf("%d\n", ans);
}