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>
#include<cstring>
#include<algorithm>
#include<unordered_map>
using namespace std;
const int N=200010,mod=998244353;
int n,m,ans,c[N],dp[N],f[N];
char a[N];
int Add(int a,int b){
	return a+b>=mod?a+b-mod:a+b;
}
int Minus(int a,int b){
	return a<b?a-b+mod:a-b;
}
int dfs(int now,int flag){
	if(now==m){
		if(!flag)
			return 1;
		else{
			while(now<n){
				if((c[now%m]^((now/m)&1))<a[now]-'0')
					return 1;
				else if((c[now%m]^((now/m)&1))>a[now]-'0')
					return 0;
				now++;
			}
			return 1;
		}
	}
	if(!flag&&~dp[now])
		return dp[now];
	int ret=0;
	for(int i=0;i<=(flag?a[now]-'0':1);i++){
		c[now]=i;
		ret=Add(ret,dfs(now+1,flag&&i==a[now]-'0'));
	}
	if(!flag)
		dp[now]=ret;
	return ret;
}
int main(){
	scanf("%d%s",&n,a);
	for(int i=1;i<=n;i++)
		if(n%i==0&&((n/i)&1)){
			memset(dp,-1,sizeof(dp));
			m=i;
			int tmp=mod-f[i];
			f[i]=Add(f[i],dfs(0,1));
			for(int j=3;i*j<=n;j+=2)
				f[i*j]=Minus(f[i*j],f[i]);
			ans=(ans+2ll*i*f[i])%mod;
		}
	printf("%d\n",ans);
	return 0;
}