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>
#define Maxn 200000
#define Mod 998244353
char s[Maxn+5];
int a[Maxn+5];
int n;
int sum[Maxn+5];
int f[Maxn+5];
int check(int x){
	int now=1,f=0;
	for(int i=1;i<=n;i++){
		if((s[now]^48^f)>(s[i]^48)){
			return 0;
		}
		else if((s[now]^48^f)<(s[i]^48)){
			return 1;
		}
		else{
			now++;
			if(now>x){
				now=1;
				f^=1;
			}
		}
	}
	return 1;
}
int get(int x){
	if((n/x)&1){
		return (sum[x]+check(x))%Mod;
	}
	return 0;
}
int main(){
	scanf("%d",&n);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++){
		a[i]=(s[i]^48);
		sum[i]=((sum[i-1]<<1)+a[i])%Mod;
	}
	for(int i=1;i*i<=n;i++){
		if(n%i==0){
			f[i]=get(i);
		}
		if(i*i!=n&&n%(n/i)==0){
			f[n/i]=get(n/i);
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=i+i;j<=n;j+=i){
			f[j]=(f[j]-f[i]+Mod)%Mod;
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		if(n%i==0){
			ans+=2ll*f[i]%Mod*i%Mod;
			ans%=Mod;
		}
	}
	printf("%d\n",ans);
	return 0;
}
