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 int long long
const int MOD=998244353;

int N,ans;
string x;
struct p{int n,s;};
vector<p> v;

int pw(int n,int k){
  int res=1;
  while(k){
    if(k&1)res*=n;res%=MOD;
    n*=n;n%=MOD;
    k>>=1;
  }
  return res;
}

int f(int a){//長さaのループがX以下に何種類存在するか
  int res=0;
  for(int i=0;i<a;i++) if(x[i]=='1') res+=pw(2,a-i-1),res%=MOD;//桁DPっぽいところ
  //最初a文字がXと完全に一致しているやつがx以下かどうかの判定
  string s,A=x.substr(0,a);
  while(s.size()<N)s+=A;
  for(int i=0;i<s.size();i++) if((i/a)&1) s[i]=(s[i]^'1')+'0';
  //sがx以下かどうか判定する
  for(int i=a;i<N;i++){
    if(x[i]^s[i]) {res+=s[i]^'1';break;}
    res+=(i==N-1);
  }
  return res+(a==N);
}

signed main(){
  cin>>N>>x;
  for(int i=1;i<=N;i++) if(N%i==0&&(N/i)&1) v.push_back({i,f(i)*2});
  for(int i=0;i<v.size();i++){
    ans+=v[i].s*v[i].n;ans%=MOD;
    for(int j=i+1;j<v.size();j++) if(v[j].n%v[i].n==0) v[j].s-=v[i].s;
  }
  if(ans<0)ans+=MOD;
  cout<<ans<<endl;
}
