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;
const int mod=998244353;
inline int add(int a,int b)
{
	if((a+=b)>=mod)a-=mod;
	return a;
}
inline int mult(int a,int b)
{
	long long t=1ll*a*b;
	if(t>=mod)t%=mod;
	return t;
}
inline int dec(int a,int b)
{
	if((a-=b)<0)a+=mod;
	return a;
}
int n,f[400010],ans;
char s[200010],x[400010],t[400010];
vector<int> v;
int main()
{
	scanf("%d%s",&n,s+1);
	for(int i=1;i<=n;i++)x[i]=s[i];
	for(int i=n+1;i<=n*2;i++)x[i]=(x[i-n]^1);
	for(int i=1;i<=2*n;i++)
		if((2*n)%i==0&&((2*n/i)&1)&&((i&1)==0))v.push_back(i);
	for(int i=0;i<v.size();i++)
	{
		int len=v[i];
		for(int j=1;j<=len/2;j++)t[j]=x[j],f[len]=add(mult(f[len],2),x[j]-'0');
		f[len]=add(f[len],1);
		for(int j=len/2+1;j<=len;j++)t[j]=(x[j-len/2]^1);
		for(int j=len+1;j<=2*n;j++)t[j]=t[j-len];
		if(strcmp(x+1,t+1)==-1)f[len]=dec(f[len],1);
		for(int j=0;j<i;j++)
			if(len%v[j]==0)f[len]=dec(f[len],f[v[j]]);
		ans=add(ans,mult(len,f[len]));
	}
	printf("%d\n",ans);
	return 0;
}