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>
#define rep(X,A,B) for(int X=A;X<=B;X++)
#define tep(X,A,B) for(int X=A;X>=B;X--)
#define LL long long
#define DB double
const int N=200010;
const int M=12000010;
const LL MOD=998244353;
using namespace std;

int n;
char s[N];
int b[N],d[N],cnt=0;
LL c[N];

void READ(){
	scanf("%d",&n);
	scanf("%s",s+1);
	rep(i,1,n)b[i]=s[i]-'0';
}

void GETK(){
	for(int i=1;i*i<=n;i++){
		if(n%i)continue;
		if(i&1)d[++cnt]=n/i;
		if(i*i==n)continue;
		if((n/i)&1)d[++cnt]=i;	
	}
	sort(d+1,d+cnt+1);
}

LL GET(int x){
	LL now=0;
	rep(i,1,x){
		now=now*2%MOD+b[i];
		now%=MOD;
	}
	int pos=0,flg=1,val;
	rep(i,1,n){
		pos=pos%x+1;
		if(pos==1)flg^=1;
		val=b[pos]^flg;
		if(val<b[i])return (now+1)%MOD;
		if(val>b[i])return now;
	}
	return (now+1)%MOD;
}

void SOLVE(){
	LL ans=0;
	rep(i,1,cnt){
		c[i]=GET(d[i]);
		rep(j,1,i-1)if(d[i]%d[j]==0)c[i]=(c[i]-c[j]+MOD)%MOD;
		ans+=c[i]*d[i]%MOD*2%MOD;ans%=MOD;
	}
	printf("%lld\n",ans);
}

int main(){
	READ();
	GETK();
	SOLVE();
	return 0;
}