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<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int mod=998244353;
int n;
char s[200005];
int a[200005],b[400005];
long long f[400005];
void Solve(int v)
{
	long long res=0;
	for(int i=1;i<=v/2;i++) res=((res<<1)+a[i])%mod;
	for(int i=1,op=0;i<=n;i+=v/2,op^=1)
	for(int j=i;j<=i+v/2-1;j++) b[j]=a[j-i+1]^op;
	bool flag=1;
	for(int i=1;i<=n;i++)
	{
		if(b[i]>a[i]) flag=0;
		else if(b[i]<a[i]) break;
	}
	if(flag) res=(res+1)%mod;
	f[v]=res;
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) a[i]=s[i]-'0';
	for(int i=2;i<=2*n;i++) if(!(2*n%i)&&n%i) Solve(i);
	for(int i=2;i<=2*n;i++)
	if(!(2*n%i)&&n%i)
	for(int j=2*i;j<=2*n;j+=i)
	if(!(2*n%j)&&n%j) f[j]=(f[j]-f[i]+mod)%mod;
	long long ans=0;
	for(int i=1;i<=2*n;i++) ans=(ans+i*f[i]%mod)%mod;
	printf("%lld\n",ans);
	return 0;
}