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;
const int N=200025,inf=0x3f3f3f3f,p=998244353;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,ans,f[N];
char ch[N];
void check(int &x,int y){x=(x+y<p) ? (x+y) : (x+y-p);}
int main(){
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
	n=read();
	scanf(" %s",ch);
	for (int i=1;i<=n;i+=2)
	if (n%i==0){
		m=n/i;
		for (int j=0;j<m;j++)
		f[i]=(f[i]*2+ch[j]-'0')%p;
		int fl=1;
		for (int j=m;j<n;j++)
		if (ch[j]!=(ch[j%m]^(j/m&1))){fl=(ch[j]>(ch[j%m]^(j/m&1)));break;}
		check(f[i],fl);
	}
	for (int i=n;i;i--){
		for (int j=i+i;j<=n;j+=i) check(f[i],p-f[j]);
		ans=(ans+(ll)n/i*2*f[i])%p;
	}
	printf("%d\n",ans);
	return 0;
}