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>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
#define pb push_back
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 200010;
int n, a[N], ans, mu[N], tx[N];
char s[N];
bool p[N];

int calc(int x) {
	int res = 0, j = 1;
	rep(i, 1, x) res = (res * 2 + a[i]) % mod;
	rep(i, 1, n) {
		tx[i] = (j <= x ? a[j] : (a[j - x] ^ 1));
		++j; if(j > x + x) j -= x + x;
	}
	rep(i, 1, n) if(a[i] != tx[i]) return tx[i] < a[i] ? (res + 1) % mod : res;
	return (res + 1) % mod;
}
int main() {
	scanf("%d%s", &n, s + 1); rep(i, 1, n) a[i] = s[i] - '0';
	rep(i, 1, n) if(n % i == 0 && (n / i) % 2) p[i] = 1;
	ans = 1ll * calc(n) * 2 * n % mod;
	per(i, n - 1, 1) if(p[i]) {
		int w = n + n;
		for(int j = i; j <= n; j += i) if(p[j]) w = (w + mu[j]) % mod;
		mu[i] = (i + i - w + mod) % mod;
		ans = (ans + 1ll * calc(i) * mu[i] % mod) % mod;
	}
	printf("%d\n", ans);
	return 0;
}