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 <iostream>
#include <string>
#include <vector>
using namespace std;

typedef long long ll;
const ll MOD = 998244353;

int main(){
	ll n;
	string x;
	cin >> n >> x;
	
	ll e=0,m=n,t=1;
	while(m%2 == 0){
		e ++;
		m /= 2;
		t *= 2;
	}
	vector<ll> div;
	for(int i = 1 ; i <= n ; i ++){
		if(m%i == 0)div.push_back(i);
	}
	
	ll cnt[200010];
	for(int i = 0 ; i < div.size() ; i ++){
		ll d = div[i];
		cnt[d] = 0;
		for(int i = 0 ; i < d*t ; i ++){
			cnt[d] *= 2;
			cnt[d] %= MOD;
			cnt[d] += x[i]-'0';
		}
		string y = "";
		for(int i = 0 ; i < d*t ; i ++){
			y += x[i];
		}
		for(int i = 0 ; i < d*t ; i ++){
			y += '0'+(1-(x[i]-'0'));
		}
		bool b = true;
		for(int i = 0 ; i < x.size () ; i ++){
			if(x[i] < y[i%(2*d*t)]){
				b = false;
				break;
			}
			if(x[i] > y[i%(2*d*t)]){
				b = true;
				break;
			}
		}
		if(b)cnt[d] ++;
		for(int j = 0 ; j < i ; j ++){
			if(d%div[j] == 0)cnt[d] += MOD-cnt[div[j]];
		}
		cnt[d] %= MOD;
	}
	
	/*for(int i = 0 ; i < div.size() ; i ++){
		cout << div[i] << " " << cnt[div[i]] << endl;
	}*/
	
	ll ret = 0;
	/*for(int i = 0 ; i < x.size() ; i ++){
		ret *= 2;
		if(x[i] == '1')ret ++;
		ret %= MOD;
	}
	ret ++;
	ret *= 2*n;
	cout << ret << endl;
	ret %= MOD;*/
	for(int i = 0 ; i < div.size() ; i ++){
		ret += cnt[div[i]]*div[i]*2*t;
		ret %= MOD;
	}
	cout << ret%MOD << endl;
}