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
#define int long long
using namespace std;
const int N=4*1e5+100;
int n,a[N],b[N],cnt[N];
char s[N];
signed main()
{
	scanf("%lld",&n);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++) a[i]=s[i]-'0';
	for (int i=1;i<=2*n;i++)
	{
		if ((2*n)%i!=0 || n%i==0) continue;
		int num=0;
		for (int j=1;j<=i/2;j++) b[j]=a[j],num=(num*2+a[j])%mod;
		cnt[i]=(cnt[i]+num)%mod;
		for (int j=i/2+1;j<=n;j+=i/2)
		{
			for (int k=j;k<j+i/2;k++)
			  b[k]=b[k-i/2]^1;
		}
		bool bl=1;
		for (int j=1;j<=n;j++)
		{
			if (b[j]>a[j])
			{
				bl=0;
				break;
			}
			if (a[j]>b[j]) break;
		}
		if (bl) cnt[i]=(cnt[i]+1)%mod;
		for (int j=i+i;j<=2*n;j+=i)
		{
			if ((2*n)%j!=0 || n%j==0) continue;
			cnt[j]=(cnt[j]-cnt[i]+mod)%mod;
		}
	}
	int ans=0;
	for (int i=1;i<=2*n;i++) ans=(ans+i*cnt[i]%mod)%mod;
	printf("%lld\n",ans);
}