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 re register
#define ull unsigned ll
using namespace std;
inline int read(){
	int s=0,t=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')t=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*t;
}
const int N=4e5+5;
const int P=998244353;
char s[N];
int n,Ans,a[N],b[N],f[N];
int Mod(int x){return x>=P?x-P:x;}
void Solve(int x){
	int res=0,flag=1;
	for(int i=1;i<=x/2;i++)res=Mod(2*res%P+a[i]);
	for(int k=1,t=0;k<=n;k+=x/2,t++)
	for(int i=k;i<k+x/2;i++)b[i]=a[i-k+1]^(t&1);
	for(int i=1;i<=n;i++){
		if(b[i]>a[i]){flag=0;break;}
		if(b[i]<a[i])break;
	}f[x]=Mod(res+flag);
}
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n=read(),scanf("%s",s+1);
	for(int i=1;i<=n;i++)a[i]=s[i]-'0';
	for(int i=1;i<=2*n;i++)if(n%i&&2*n%i==0)Solve(i);
	for(int i=1;i<=2*n;i++)if(n%i&&2*n%i==0)
	for(int j=2*i;j<=2*n;j+=i)if(n%j&&2*n%j==0)f[j]=Mod(f[j]+P-f[i]);
	for(int i=1;i<=2*n;i++)Ans=Mod(Ans+1ll*i*f[i]%P);
	printf("%d",Ans);
	return 0;
}