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>
const int N=200005,P=998244353;
int n,f[N],p[N],a;char c[N];
inline int ck(int x)
{
	for(int i=1,j=1,w=0;i<=n;++i,j==x?(j=1,w^=1):++j)
		if(c[i]<(c[j]^w))return 0;
		else if(c[i]>(c[j]^w))return 1;
	return 1;
}
int main()
{
	scanf("%d%s",&n,c+1);
	p[0]=1;for(int i=1;i<=n;++i)p[i]=p[i-1]*2%P;
	for(int d=1;d<=n;++d)if(!(n%d)&&n/d&1)
	{
		for(int i=1;i<=d;++i)
		if(c[i]=='1')(f[d]+=p[d-i])%=P;
		(f[d]+=ck(d))%=P;
		for(int g=1;g<d;++g)if(!(d%g)&&n/g&1)
			(f[d]+=P-f[g])%=P;
		a=(2ll*d*f[d]+a)%P;
	}
	return printf("%d",a),0;
}