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 mod=998244353;
const int N=2e5+2;
int num[N];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l;
	long long ans=0;
	string s;
	cin>>n>>s;
	for(i=1;i<=n;i++){
		if((n%i)||!(n%(2*i))){
			continue;
		}
		string t="",rev="",t1="";
		k=0;
		for(j=0;j<i;j++){
			t+=s[j];
			rev+=(s[j]^('1'^'0'));
			k=(k*2+(int)(s[j]-'0'));
			if(k>=mod){
				k-=mod;
			}
		}
		//cout<<i<<' '<<k<<endl;
		num[i]+=k;
		for(j=0;j<n/i;j++){
			if(!(j&1)){
				t1+=t;
			}
			else{
				t1+=rev;
			}
		}
		if(t1<=s){
			num[i]++;
		}
		if(num[i]>=mod){
			num[i]-=mod;
		}
		ans+=2ll*num[i]*i;
	//	cout<<i<<' '<<num[i]<<endl;
		ans%=mod;
		for(j=3*i;j<=n;j+=2*i){
			num[j]-=num[i];
			if(num[j]<0){
				num[j]+=mod;
			}
		}
	}
	cout<<ans;
}