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 <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll MOD=998244353;

ll N;
string S;

ll dp[200020][2];

void add(ll &a,ll b){
  a=(a+b)%MOD;
}

ll mod_pow(ll x,ll n){
  x%=MOD;
  ll res=1;
  while(n>0){
    if(n&1) res=res*x%MOD;
    x=x*x%MOD;
    n/=2;
  }
  return res;
}

int main(){
  cin>>N;
  cin>>S;

  dp[0][0]=1;
  for(int i=0;i<N;i++){
    if(S[i]=='0'){
      add(dp[i+1][0],dp[i][0]);
      add(dp[i+1][1],dp[i][1]*2%MOD);
    }else{
      add(dp[i+1][0],dp[i][0]);
      add(dp[i+1][1],dp[i][1]*2%MOD);
      add(dp[i+1][1],dp[i][0]);
    }
  }

  vector<pair<int,ll>> v;

  for(int i=1;i<=N;i++){
    if(N%i>0) continue;
    int x=N/i;
    if(x%2==0) continue;

    string s="";
    for(int j=0;j<i;j++) s+=S[j];
    string t="";
    for(int j=0;j<i;j++){
      if(s[j]=='0') t+="1";
      else t+="0";
    }
    string r="";
    for(int j=1;j<=x;j++){
      if(j%2) r+=s;
      else r+=t;
    }

    if(S>=r) v.push_back(mkp(i,dp[i][1]+1));
    else v.push_back(mkp(i,dp[i][1]));
  }

  ll ans=0;
  for(int i=0;i<v.size();i++){
      int now=v[i].first;
      add(ans,v[i].second*2*now%MOD);
      for(int j=i+1;j<v.size();j++){
          if(v[j].first%now==0) add(v[j].second,MOD-v[i].second);
      }
  }

  cout<<ans<<endl;

  return 0;
}
