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;
const int N=2e5+5;
const int mod=998244353;
int n,f[N],ans;char str[N],ss[N];
int sum(int m){
	int res=0;
	for(int i=1;i<=m;++i)res=(res<<1|str[i]-'0')%mod;
	return res;
}
int check(int m){
	for(int i=1;i<=m;++i)ss[i]=str[i];
	for(int i=m+1;i<=n;++i)ss[i]=ss[i-m]^1;
	for(int i=1;i<=n;++i)
		if(ss[i]!=str[i])
			return ss[i]<str[i];
	return 1;
}
int main(){
	scanf("%d",&n);
	scanf("%s",str+1);
	for(int i=1;i<=n;++i)
		if(n%i==0&&n/i&1)
			f[i]=sum(i)+check(i);
	for(int i=1;i<=n;++i)
		if(n%i==0)
			for(int j=1;j<i;++j)
				if(i%j==0)
					f[i]=(f[i]-f[j]+mod)%mod;
	for(int i=1;i<=n;++i)
		ans=(ans+2ll*f[i]*i)%mod;
	printf("%d\n",ans);
	return 0;
}
