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<vector>
using namespace std;
int N;
string s;
long mod=998244353;
long sum[2<<17];
long dp[2][2<<17];
int main()
{
	cin>>N>>s;
	dp[0][0]=1;
	for(int i=0;i<=N;i++)
	{
		if(i!=0)
		{
			if(N%i==0&&N/i%2==1)
			{
				sum[i]=dp[1][i];
				bool flag=true;
				for(int k=0;k<N;k++)
				{
					int L=s[k]-'0';
					int R=s[k%i]-'0'^(k/i%2==1);
					if(L!=R)
					{
						if(L<R)flag=false;
						break;
					}
				}
				if(flag)(sum[i]+=dp[0][i])%=mod;
			}
		}
		if(i==N)break;
		for(int j=0;j<2;j++)
		{
			int lim=j?1:s[i]-'0';
			for(int k=0;k<=lim;k++)
			{
				(dp[j||k<lim][i+1]+=dp[j][i])%=mod;
			}
		}
	}
	long ans=0;
	for(int i=1;i<=N;i++)
	{
		if(sum[i]==0)continue;
		for(int j=i*2;j<=N;j+=i)
		{
			if(N%j==0&&N/j%2==1)(sum[j]+=mod-sum[i])%=mod;
		}
		(ans+=sum[i]*i*2)%=mod;
	}
	cout<<ans<<endl;
}
