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;
const int N=2e5+10;
const int mod=998244353;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,f[N],ans=0,bin[N];
char s[N];

int main() {
	cin>>n;scanf("%s",s+1);reverse(s+1,s+n+1);
	bin[0]=1;for(int i=1;i<=n;i++) bin[i]=(bin[i-1]<<1)%mod;
	for(int i=1;i<=n;i++)
		if(n%i==0&&((n/i)&1)) {
			int sum=0;
			for(int j=n;j>=n-i+1;j--)
				if(s[j]=='1') sum=(sum+bin[j-(n-i)-1])%mod;
			int fl=0,fl2=0;
			for(int j=n-i;j&&!fl2;j-=i) {
				fl^=1;
				for(int k=j,l=n;k>=j-i+1;k--,l--) {
					int b=(s[l]-'0')^fl;
					if(b<s[k]-'0') { sum=(sum+1)%mod;fl2=1;break; }
					else if(b>s[k]-'0') { fl2=1;break; }
				}
			}
			if(!fl2) sum=(sum+1)%mod;
			f[i]=sum;
		}
	for(int i=1;i<=n;i++)
		if(n%i==0&&((n/i)&1)) 
			for(int j=i+i;j<=n;j+=i)
				if(n%j==0&&((n/i)&1)) f[j]=(f[j]-f[i]+mod)%mod;
	for(int i=1;i<=n;i++)
		if(n%i==0&&((n/i)&1)) ans=(ans+2ll*f[i]*i)%mod;
	cout<<ans;
	return 0;
}
