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;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = 998244353ll;

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

vector<int> v;

int main(void){
	int n,m,p,q,i,j,k;
	ll tmp,ans=0ll;
	bool flag;
	string s,t;
	cin >> n;
	cin >> s;
	t = s;
	for(i=1; i*i<=n; ++i){
		if(n%i){
			continue;
		}
		if(i%2){
			v.push_back(i);
		}
		if(n/i==i){
			continue;
		}
		if((n/i)%2){
			v.push_back(n/i);
		}
	}
	sort(v.begin(),v.end());
	m = v.l_ength();
	vector<ll> w(m,0ll);
	for(i=0; i<m; ++i){
		p = v[i];
		q = n/v[i];
		add_mod(w[i],q*2);
		for(j=0; j<i; ++j){
			if(v[i]%v[j]){
				continue;
			}
			add_mod(w[i],MOD-w[j]);
		}
		tmp = 0ll;
		for(k=0; k<q; ++k){
			mul_mod(tmp,2ll);
			add_mod(tmp,s[k]-'0');
		}
		for(j=0; j<p; ++j){
			for(k=0; k<q; ++k){
				t[j*q+k] = ((j%2)?('0'+'1'-s[k]):s[k]);
			}
		}
		flag = true;
		for(j=0; j<n; ++j){
			if(s[j]<t[j]){
				flag = false;
				break;
			}else if(s[j]>t[j]){
				break;
			}
		}
		if(flag){
			add_mod(tmp,1ll);
		}
		mul_mod(tmp,w[i]);
		add_mod(ans,tmp);
	}
	cout << ans << endl;
	return 0;
}
