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<iostream>
#include<cstdio>
const int N=2e5+5;
typedef long long ll;
const ll mod=998244353;
int n;
char s[N],t[N];
ll f[N];
bool cmp() {
    for(int i=0;i<n;i++) {
        if(s[i]==t[i]) continue;
        return s[i]>t[i];
    }
    return true;
}
int main() {
    scanf("%d",&n);
    scanf("%s",s);
    ll ans=0;
    for(int i=0;i<n;i++) {
        ans=(ans*2+s[i]-'0')%mod;
    }
    f[n]=(ans+1)%mod;
    for(int i=3;i<=n;i+=2) if(n%i==0) {
        ll tot=0;
        int len=n/i;
        for(int j=0;j<n/i;j++) tot=(tot*2+s[j]-'0')%mod;
        f[len]=tot;
        for(int j=0;j<len;j++) {
            t[j]=s[j];
        }
        for(int j=len;j<n;j++) t[j]=((t[j-len]-'0')^1)+'0';
        if(cmp()) f[len]=(f[len]+1)%mod;
    }ans=0;
    for(int i=1;i<=n;i++) {
        if(n%i==0&&(n/i)&1) {
            for(int k=i+i;k<=n;k+=i) f[k]=(f[k]-f[i]+mod)%mod;
            ans=(ans+2*i*f[i])%mod;
        }
    }
    printf("%lld\n",ans);
    return 0;
}
