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 mod 998244353
using namespace std;
typedef long long int ll;
const int maxn=2E5+5;
int n;
string str;
ll ans,f[maxn];
inline ll solve(int x)
{
	string base1,base2,s;
	ll g=0;
	for(int i=0;i<x;++i)
	{
		base1+=str[i],base2+='1'-(str[i]-'0');
		g=(g*2+str[i]-'0')%mod;
	}
	s=base1;
	++g;
	for(int i=0;i<(n/x-1)/2;++i)
		s+=base2+base1;
	g-=s>str;
	return g;
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>str;
	for(int i=1;i<=n;++i)
		if(n%i==0&&(n/i)%2==1)
			f[i]=solve(i);
	for(int i=1;i<=n;++i)
		for(int j=i*2;j<=n;j+=i)
			f[j]=(f[j]-f[i]+mod)%mod;
	for(int i=1;i<=n;++i)
		if(n%i==0)
			ans=(ans+f[i]*2*i)%mod;
	ans=(ans%mod+mod)%mod;
	cout<<ans<<endl;
	return 0;
}
