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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ld long double
#define al(a) (a).begin(),(a).end()
#define mk make_pair
#define check cout<<"?"<<endl;

ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;

int main(){
    ll n,ans=0,num=0,l=0; 
    string x; cin>>n>>x;
    vector<ll> c(n+1,0),p;
    for(ll i=1;i<=n;i++){
        if(n%i==0 && n/i%2) p.push_back(i);
    }
    ll N=p.size();
    ll k,t1,t2;
    rep(i,N){
        while(l<p[i]){
            num=(num*2+x[l]-'0')%mod;
            l++;
        }
        k=num+1;
        for(int j=p[i];j<n;j++){
            t1=x[j]-'0';
            t2=(x[j%p[i]]-'0'+j/p[i])%2;
            if(t1!=t2){
                if(t2) k--;
                break;
            }
        }
        rep(j,i)if(p[i]%p[j]==0 && p[i]/p[j]%2)
        k=(k-c[p[j]])%mod;
        c[p[i]]=k;
        ans=(ans+k*2*p[i]%mod)%mod;
        //cout<<p[i]*2<<" "<<k<<endl;
    }
    cout<<(mod+ans)%mod<<endl;
}
