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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const long long mod=998244353;

int main()
{
  int n;
  static char s[200010];
  scanf("%d%s",&n,s);
  vector<bool> X(n);
  for(int i=0;i<n;i++){
    X[i]=(s[i]=='1');
  }
  vector<int> ds;
  for(int t=1;t*t<=n;t++){
    if(n%t==0){
      ds.push_back(t);
      if(n/t!=t){
	ds.push_back(n/t);
      }
    }
  }
  sort(ds.begin(),ds.end());
  int dn=ds.size();
  vector<long long> C(dn,0ll);
  long long ans=0ll;
  for(int t=0;t<dn;t++){
    int d=ds[t];
    if(d%2==0)
      continue;
    long long c=0;
    for(int i=0;i<n/d;i++){
      c=(2*c+X[i])%mod;
    }
    bool F=true;
    for(int i=n/d;i<n;i++){
      int x=(i/(n/d))%2==1?1-X[i%(n/d)]:X[i%(n/d)];
      if(X[i]!=x){
	F=(X[i]>x);
	break;
      }
    }
    if(F){
      c=(c+1)%mod;
    }
    C[t]=2*(n/d);
    for(int tt=0;tt<t;tt++){
      if(ds[t]%ds[tt]==0){
	C[t]=(C[t]+mod-C[tt])%mod;
      }
    }
    //printf("%d %lld %lld %lld\n",d,C[t],C[t]-mod,c);
    ans=(ans+C[t]*c%mod)%mod;
  }
  printf("%lld\n",ans);
  return 0;
}
