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;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
#define MOD 998244353
using vi = vector<ll>; // intの1次元の型に vi という別名をつける
using vvi = vector<vi>; // intの2次元の型に vvi という別名をつける

ll n,k,x,y;

int main(){
  ll ans=0;
  cin >> n;
  string s;cin>>s;

  
  REP(i,n){
    ans*=2LL;
    ans+=(ll)(s[i]-48);
    ans%=MOD;
  }
  ans++;
  ans*=2LL*n;
  ans%=MOD;
  vi pt(n/3+1,0);


  for(ll i=1;i<=n/3;i++){
    if (n%(i*2)==i){

      ll flg=0;
      for(int j=i;j<n;j++){
        
        if ((j/i)%2==0){
          if (s[j]<s[j%i]){
            flg=1;
            break;
          }
          if (s[j]>s[j%i]){
            flg=0;
            break;
          }
        }else{
          if (s[j]<(49-s[j%i]+48)){
            flg=1;
            break;
          }
          if (s[j]>(49-s[j%i]+48)){
            flg=0;
            break;
          }
        }
      //ある場合はflg0 ない場合はflg1
      }

        ll sm=0LL;
        REP(j,i){
          sm*=2LL;
          sm+=(ll)(s[j]-48);
          sm%=MOD;
        }
        sm++;
        sm=(sm-flg+MOD)%MOD;

        
        REP(jj,i){
          if (jj==0){
            continue;
          }
          //if ((jj*jj)>=i)break;
          if (i%jj==0){
            if ((i/jj)%2==1){
              sm=(sm-pt[jj]+MOD)%MOD;
            }
          }
        }
        

        pt[i]=sm;
        ans-=(sm*(2LL*n));
        ans+=sm*(2LL*i);
        ans%=MOD;
        ans=(ans+MOD)%MOD;
        //cout<<i<<endl;
    }
  }




  cout<<ans<<endl;
  return 0;
}