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<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int Mod=998244353;

int n,nn; char S[200010];
int ans[400010];
int solve(int x)
{
	x>>=1;
	int res=0;
	for (int i=1;i<=x;i++) res=(res*2+S[i]-'0')%Mod;
	bool flg=false;  //greater or not
	for (int i=x+1,j=1,k=1;i<=n;i++)
	{
		char t=S[j]^k;
		if (t!=S[i])
		{
			flg=S[i]<t; break;
		}
		if (j==x) j=1,k^=1;
		else j++;
	}
	
	//if (S[i]!=S[(i-1)%x+1]){flg=S[i]<S[(i-1)%x+1];break;}
	if (!flg) res=(res+1)%Mod;
	//printf("solve %d %d\n",x*2,res);
	return res;
}
int main()
{
	scanf("%d%s",&n,S+1); nn=n+n;
	for (int i=nn;i;i-=2)
		if (nn%i==0&&n%i==(i>>1))
			ans[i]=solve(i);
	//for (int i=1;i<=nn;i++) printf("ans %d: %d\n",i,ans[i]);
	for (int i=2;i<=nn;i+=2)
		if (nn%i==0&&n%i==(i>>1))
			for (int j=i+i;j<=nn;j+=i)
				if (nn%j==0&&n%j==(j>>1))
					ans[j]=(ans[j]-ans[i])%Mod;
	int res=0;
	for (int i=2;i<=nn;i+=2) res=(res+1LL*i*ans[i])%Mod;
	printf("%d\n",(res+Mod)%Mod);
	return 0;
}
