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>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MD = 998244353;
const int MX = 200005;
int n;
long long m[MX], ans, sm, f[MX], mp[MX];
string x;
int gcd(int x, int y){
	if(x < y) swap(x, y);
	while(y){
		x %= y;
		swap(x,y);
	}
	return x;
}
void setmap(int t){
	int u = t;
	for(; u < n && x[u-t] != x[u]; u++);
	mp[t] = f[t] + (u == n || x[u] == '1');
}
int main(){
	cin >> n >> x;
	rep(i,n) f[i+1] = (f[i]*2 + x[i] - '0') % MD;
	for(int i = 1; i*i <= n; i++) if(!(n%i)){
		setmap(i);
		if(i*i != n) setmap(n/i);
	}
	for(int i = 1; i < n; i++){
		int t = gcd(i,n);
		if(i/t % 2 == 0){
			m[i] = (m[i] + mp[t]) % MD;
			sm = (sm + m[i]) % MD;
			ans = (ans + m[i] * i) % MD;
			for(int j = i*2; j < n; j += i) m[j] = (m[j] - m[i] + MD) % MD;
		}
	}
	printf("%lld\n", (ans + (f[n]+1 - sm + MD) * 2*n) % MD);
}