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;
#define pb push_back
#define Fast_IO ios::sync_with_stdio(false);
#define int long long
inline int read()
{
	char ch=getchar(); int nega=1; while(!isdigit(ch)) {if(ch=='-') nega=-1; ch=getchar();}
	int ans=0; while(isdigit(ch)) {ans=ans*10+ch-48;ch=getchar();}
	if(nega==-1) return -ans;
	return ans;
}
#define mod 998244353
#define N 200005
int n;
string s;
map<int,int> a;
int work(int len)
{
	int v=0;
	for(int i=0;i<len;i++)
	{
		v=(v*2+s[i]-'0')%mod;
	}
	int ok=1;
	for(int i=len;i<n;i++)
	{
		int x=s[i%len]-'0';
		if(i/len%2) x^=1;
		if(x==s[i]-'0') continue;
		if(x>s[i]-'0') ok=0;
		break;
	}
	v=(v+ok)%mod;
	int ans=0;
	map<int,int>::iterator i;
	for(i=a.begin();i!=a.end();++i)
	{
		if(len%(i->first)==0) v=(v-i->second+mod)%mod;
	}
	a[len]=v;
	return (ans+2*v*len)%mod;
}
signed main()
{
#ifdef __LOCAL__
	freopen("in.txt","r",stdin);
#endif
	cin>>n;
	cin>>s;
	int ans=0;
	for(int i=1;i<=n;i++)
	{
//		cout<<i<<endl;
		if(n%i==0&&n/i%2==1)
		{
			ans=(ans+work(i))%mod;
		}
	}
	cout<<ans<<endl;
#ifdef __LOCAL__
	printf("Time Used : %d\n",clock());
#endif
	return 0;
}

