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>
const int MaxN=200010,mod=998244353;
int N,pw2[MaxN],f[MaxN*2];
char X[MaxN];
int main(){
	scanf("%d%s",&N,X);
	for(int i=*pw2=1;i<=N;i++)pw2[i]=2ll*pw2[i-1]%mod;
	for(int k=1;k<=N;k+=2)if(N%k==0){
		int L=N/k;
		bool fl=1;
		for(int i=0;i<N&&fl;i++){
			if(X[i]=='1'){
				if(i<L||X[i-L]=='1'){
					f[L]=(f[L]+(i<L?pw2[L-i-1]:1))%mod;
				}
			}
			if(i>=L&&X[i]==X[i-L])fl=0;
		}
		if(fl)f[L]=(f[L]+1)%mod;
	}
	int ans=0;
	for(int i=1;i<=N;i++)if(N%i==0&&N/i%2==1&&f[i]){
		ans=(ans+2ll*f[i]*i)%mod;
		for(int j=i*2;j<=N;j+=i)f[j]=(f[j]+mod-f[i])%mod;
	}
	printf("%d\n",ans);
}