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<cstdio>
#include<cmath>
using namespace std;
const int N=200005;
const int MOD=998244353;
int n;
int a[N];
long long solve(int T)
{
	static int p[N],q[N];
	for(int i=1;i<=T/2;i++)
		p[i]=a[i],q[i]=a[i]^1;
	static int b[N];
	for(int i=1;i<=n*2;)
	{
		for(int j=1;j<=T/2;j++)
			b[i+j-1]=p[j];
		i+=T/2;
		for(int j=1;j<=T/2;j++)
			b[i+j-1]=q[j];
		i+=T/2;
	}
	long long ans=0;
	for(int i=1;i<=T/2;i++)
		ans=(ans*2+a[i])%MOD;
	ans=(ans+1)%MOD;
	for(int i=1;i<=n;i++)
		if(b[i]<a[i]) break;
		else if(b[i]>a[i])
		{
			ans=(ans-1+MOD)%MOD;
			break;
		}
	return ans;
}
long long f[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%1d",&a[i]);
	for(int i=1;i<=n*2;i++)
		if(n*2%i==0&&n%i!=0) f[i]=solve(i);
	for(int i=1;i<=n*2;i++)
		if(n*2%i==0&&n%i!=0)
			for(int j=i+i;j<=n*2;j+=i)
				if(n*2%j==0&&n%j!=0) f[j]=(f[j]-f[i]+MOD)%MOD;
	long long ans=0;
	for(int i=1;i<=n*2;i++)
		if(n*2%i==0&&n%i!=0) ans=(ans+f[i]*i%MOD)%MOD;
	printf("%lld",ans);
	return 0;
}