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 int long long
const int mod=998244353;
const int N=2e5+5;
char s[N],t[N];int n,pw[N];
bool cmp(char *a,char *b){
	for(int i=1;i<=n;++i)
		if(a[i]<b[i]) return 1;
		else if(a[i]>b[i]) return 0;
	return 1;
}
int calc(int d){
	int ans=0;
	for(int i=1;i<=d;++i)
		if(s[i]=='1') ans=(ans+pw[d-i])%mod;
	for(int st=1;st<=n;st+=d)
		for(int j=st;j<st+d;++j)
			t[j]=(((st-1)/d)&1)?(s[j-st+1]^1):(s[j-st+1]);
	//cout<<d<<endl;
	//for(int i=1;i<=n;++i)
	//	cout<<t[i];
	//cout<<endl;
	if(cmp(t,s)) ++ans;
	return ans%mod;
}
vector<int> vec;
int f[N];
signed main(){
	cin>>n>>(s+1);
	pw[0]=1;
//	for(int i=1;i<=n;++i)cout<<s[i];cout<<endl;
	for(int i=1;i<=n;++i)
		pw[i]=pw[i-1]*2ll%mod;
	for(int i=1;i<=n;++i)
		if(n%i==0 && (n/i)%2==1) vec.push_back(i);
	int ans=0;
	for(int i=0;i<vec.size();++i){
		int u=vec[i];
		f[i]=calc(u);
		for(int j=0;j<i;++j)
			if(vec[i]%vec[j]==0) f[i]=(f[i]+(mod-f[j])%mod)%mod;
		ans=(ans+2ll*f[i]*u%mod)%mod;
	}
	cout<<ans;
	return 0;
}