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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod=998244353;
#define N 200020
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,t[N],ans,a[N],ok[N];
char s[N];
bool check(int x){
	for(int i=0,p=0;i<n;i+=x,p^=1){
		for(int j=1;j<=x;++j){
			if(a[i+j]>(a[j]^p))return true;
			if(a[i+j]<(a[j]^p))return false;
		}
	}
	return true;
}
void Solve(int x){
	int tot=0;
	for(int i=1;i<=x;++i){
		tot=((tot<<1)+a[i])%mod;
	}
	++tot;
	if(!check(x))--tot;
	t[x]=(t[x]+tot)%mod;
	for(int i=x+x;i<=n;i+=x){
		if(ok[i])t[i]=(t[i]-t[x]+mod)%mod;
	}
}
int main(){
	n=read();
	scanf("%s",s+1);
	for(int i=1;i<=n;++i){
		a[i]=s[i]-'0';
	}
	for(int d=1;d<=n;++d){
		if(n%d==0&&(n/d)&1){
			ok[d]=1;
		}
	}
	for(int d=1;d<=n;++d){
		if(ok[d])Solve(d);
	}
	for(int i=1;i<=n;++i){
		ans=(ans+2LL*i%mod*t[i])%mod;
	}
	printf("%d\n",ans);
	return 0;
}

