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 <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define PB push_back
#define MP make_pair
#define PII pair<int,int>
#define FIR first
#define SEC second
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=2e5+10,mod=998244353;
int pw[N],f[N],n;
char str[N]; int S[N];
int sol(int k) {
    int tot=0;
    for(int i=0;i<k;++i) if(S[i]==1) tot=(tot+pw[k-i-1])%mod;
    int flg=1;
    for(int i=k;i<n;++i) if(S[i]!=(S[i-k]^1)) {
        if(S[i]<(S[i-k]^1)) flg=0;
        else flg=1;
        break;
    }
    return (flg+tot)%mod;
}
int main() {
    rd(n);
    scanf("%s",str);
    for(int i=0;i<n;++i) S[i]=str[i]-'0';
    pw[0]=1; for(int i=1;i<=n;++i) pw[i]=pw[i-1]*(ll)2ll%mod;
    for(int i=1;i<=n;++i) if(n%i==0&&((n/i)&1)) f[i]=sol(i);
    for(int i=1;i<=n;++i) if(f[i])
        for(int j=i*2;j<=n;j+=i) if(n%j==0&&((n/j)&1))
            f[j]=(f[j]-f[i])%mod;
    // for(int i=1;i<=n;++i) printf("%d ",f[i]); puts("");
    int ans=0;
    for(int i=1;i<=n;++i) ans=(ans+f[i]*2ll*i)%mod;
    printf("%d",(ans+mod)%mod);
    return 0;
}