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;
typedef long long ll;
#define ri register int
const int N=500005,M=998244353;
int n;
long long f[N],ans;
char c[N];
void Count(int p)
{
	for(ri i=0;i<p;++i)
		f[p]=(f[p]*2+c[i]-'0')%M;
	++f[p];
	for(ri i=p;i<n;++i)
	{
		if(((c[i%p]-'0')^(i/p&1))<c[i]-'0')
			break;
		if(((c[i%p]-'0')^(i/p&1))>c[i]-'0')
		{
			--f[p];
			break;
		}
	}
	for(ri i=1;i<p;++i)
		if(p%i==0)
			f[p]=(f[p]-f[i])%M;
}
int main()
{
	scanf("%d",&n);
	scanf("%s",c);
	for(ri i=1;i<=2*n;++i)
		if(2*n%i==0)
			if(2*n/i&1)
				Count(i/2);
	for(ri i=1;i<=n;++i)
		ans=(ans+f[i]*2*i)%M;
	cout<<(ans%M+M)%M;
}
