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>
#define int long long
#define mod 998244353
int n,ans,a[300100],p[300100],cnt,sum[300100];
char s[300100];
signed main(){
	scanf("%lld",&n);
	scanf("%s",s);
	for(int i=0;i<n;i++)
		a[i]=s[i]-'0';
	p[++cnt]=1;
	for(int i=3;i<=n;i+=2)
		if(n%i==0)p[++cnt]=i;
	for(int i=cnt;i>=1;i--){
		int len=n/p[i];
		for(int j=0;j<len;j++)
			sum[i]=(sum[i]*2+a[j])%mod;
		sum[i]++;
		for(int j=0;j<n;j++)
			if(((a[j%len])^(j/len&1))<a[j])break;
			else if(((a[j%len])^(j/len&1))>a[j]){
					--sum[i];
					break;
				}
		for(int j=i+1;j<=cnt;j++)
			if(p[j]%p[i]==0)(sum[i]-=sum[j]-mod)%=mod;
		(ans+=sum[i]*n/p[i]*2%mod)%=mod;
	}
	printf("%lld\n",ans%mod);
}