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>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int MAXN=4e5+5;
char x[MAXN];
int a[MAXN],b[MAXN];
int n;
ll cnt[MAXN],ans;
int calc(int len){
	ll res=0,cnt=0;
	for (int i=1;i<=len;i++){
		res=(res*2ll%mod+a[i])%mod;
	}
	for (int i=0;i<2*n;i+=len*2){
		for (int j=1;j<=len;j++) b[i+j]=a[j];
		for (int j=len+1;j<=2*len;j++) b[i+j]=!a[j-len];
	}
	cnt=1;
	while (cnt<=2*n&&a[cnt]==b[cnt]) cnt++;
	res=(res+(cnt>2*n||a[cnt]>b[cnt]))%mod;
	return res;
}
int main(){
	scanf("%d",&n);
	scanf("%s",x+1);
	for (int i=1;i<=n;i++){
		a[i]=(x[i]=='1');
		a[i+n]=(x[i]!='1');
	}
	for (int i=1;i<=n;i++){
		if (n%i==0&&(n/i)&1) cnt[i]=calc(i);
	}
	for (int i=1;i<=n;i++){
		for (int j=2*i;j<=n;j+=i){
			if (n%j==0&&(n/j)&1) cnt[j]=((cnt[j]-cnt[i])%mod+mod)%mod;
		}
	}
	for (int i=1;i<=n;i++) ans=(ans+2ll*i%mod*cnt[i]%mod)%mod;
	printf("%lld\n",ans);
	return 0;
}
