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<bits/stdc++.h>
using namespace std;

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}

void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++)

const int maxn = 1e6+10;
const int inf = 1e9;
const lf eps = 1e-8;
const int mod = 998244353;

char s[maxn];
int vis[maxn],f[maxn],n,t[maxn],cnt[maxn];

int calc(int x) {
    int ans=0;
    for(int i=1;i<=x;i++) ans=((ans<<1)|s[i])%mod,t[i]=s[i];
    for(int i=x+1;i<=n;i++) t[i]=t[i-x]^1;
    for(int i=1;i<=n;i++) {
        if(s[i]>t[i]) return ans+1;
        if(s[i]<t[i]) return ans;
    }return ans+1;
}

int main() {
    read(n);scanf("%s",s+1);
    for(int i=1;i<=n;i++) s[i]-='0';
    for(int i=1;i<=n;i++) if(n%i==0&&((n/i)&1)) vis[n/i]=1,f[n/i]=calc(i);
    //,printf("%d %d\n",i,f[n/i]);puts("OK");
    int ans=0;
    for(int i=n;i;i--) {
        if(!vis[i]) continue;
        for(int j=i<<1;j<=n;j+=i) if(vis[j]) f[i]=(f[i]-f[j]+mod)%mod;
        // printf("%d %d\n",i,f[i]);
        ans=(ans+1ll*n/i*2*f[i]%mod)%mod;
    }write(ans);
    return 0;
}
