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 ll long long
#define p 998244353
using namespace std;
int n,m,k;
char c[200010];
void b(int x){
	for(int i=1;i*x+x<=n;i+=2){
		for(int j=1;j<=x;j++)c[i*x+j]='1'+'0'-c[i*x+j];
	}
}
int v[200010],s[200010];
int calc(int q){
	int ret=0;
	for(int i=1;i<=q;i++)
	ret=(1ll*ret*2+(c[i]=='1'))%p;
	return ret;
}
int main(){
	cin>>n;
	scanf("%s",c+1);int ret=0;
	for(int i=1;i<=n;i++){
		ret=(1ll*ret*2+(c[i]=='1'))%p;
	}
	int ans=0,sum=0;
	for(int i=1;i<=n;i++)if(n%i==0)if((n/i)&1){//连续i个相同 
		b(i);
		for(int j=i;j+i<=n;j+=i){
			for(int t=1;t<=i;t++)if(c[t]!=c[j+t]){
				if((c[t]>c[j+t]&&((j/i)&1))||((c[t]<c[j+t]&&((j/i)&1^1)))){
					v[i]=calc(i)+1;
					goto ed;
				}
				else {
					v[i]=calc(i);
					goto ed;
				}
			}
		}
		v[i]=calc(i)+1;
		
		ed:;
		(v[i]-=s[i])%=p;
		for(int j=i;j<=n;j+=i)(s[j]+=v[i])%=p;(sum+=v[i])%=p;
		(ans+=1ll*v[i]*i%p*2%p)%=p;
		b(i);
	}
	cout<<(ans%p+p)%p;;
    return 0;
}






