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>
typedef long long ll;
const int mod=998244353;
char s[200050];
bool a[200050],b[200050];
int n,ans=0,tmp[200050],res[200050];
inline bool cmp(){
	for(int i=1;i<=n;++i)if(a[i]!=b[i])return a[i]<b[i];
	return 1;
}
int main(){
	int i,j,k;
	scanf("%d%s",&n,s+1);
	for(i=1;i<=n;++i){
		b[i]=s[i]=='1';
		tmp[i]=((tmp[i-1]<<1)+b[i])%mod;
	}
	ans=(ll)(tmp[n]+1)*n*2%mod;
	for(k=1;k<=n/3;++k)if(!(n%k)&&n/k%2==1){
		res[k]=tmp[k];
		for(i=1;i<=k;++i)a[i]=b[i];
		for(;i<=n;i+=k)
			for(j=0;j<k;++j)a[i+j]=a[i+j-k]^1;
		if(cmp())res[k]=(res[k]+1)%mod;
		if(k>1)for(i=1;i*i<=k;++i)if(!(k%i)){
			res[k]=(res[k]-res[i]+mod)%mod;
			if(i*i!=k&&i!=1)res[k]=(res[k]-res[k/i]+mod)%mod;
		} 
		ans=(ans-(ll)res[k]*(n-k<<1)%mod+mod)%mod;
	}
	printf("%d\n",ans);
	return 0;
}