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>
using namespace std;

const int maxn=200010,mod=998244353;
int n,a[maxn],tot,b[maxn],c[maxn],ans,d[maxn];
char s[maxn];

int inc(int a,int b)
{
	a+=b;
	if (a>=mod)
		return a-mod;
	return a;
}

int mul(int a,int b)
{
	return 1ll*a*b-1ll*a*b/mod*mod;
}

int main()
{
	int i,j,k;
	scanf("%d%s",&n,s+1);
	for (i=1;i<=n;i++)
		a[i]=(s[i]^48);
	for (k=1;k<=n;k++)
		if (n%k==0&&((n/k)&1))
		{
			b[++tot]=(k<<1);
			for (i=1;i<=k;i++)
				c[tot]=((c[tot]<<1)+a[i])%mod;
			c[tot]++;
			for (i=1;i<=k;i++)
				d[i]=a[i];
			for (i=k+1;i<=n;i++)
				d[i]=d[i-k]^1;
			for (i=k+1;i<=n;i++)
				if (a[i]^d[i])
				{
					c[tot]-=(a[i]<d[i]);
					break;
				}
		}
	for (i=1;i<=tot;i++)
		for (j=1;j<=i-1;j++)
			if (b[i]%b[j]==0)
				c[i]=inc(c[i],mod-c[j]);
	for (i=1;i<=tot;i++)
		ans=inc(ans,mul(b[i],c[i]));
	printf("%d\n",ans);
	return 0;
}