Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

3
111

Sample Output 1

40

For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

Sample Input 2

6
110101

Sample Output 2

616

Sample Input 3

30
001110011011011101010111011100

Sample Output 3

549320998

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include"stdio.h"
#include"stdlib.h"
#include"string.h"
#include"math.h"

int main(void){
  int N;
  char tmp[1000];
  int decimal=0,i,j=1000000000,k;
  int len1,len2;
  int base=1;
  int tmp2;
  int plus =0;
  int count,all_count=0;
  char* X = (char*) calloc(100,sizeof(char));
  scanf("%d",&N);
  scanf("%s",X);
  
  
  for(i=0;X[i]=='0';i++){
    len1=i;
    plus = 1;
  }
  for(i=0;X[i]!='\0';i++){
    len2=i;
  }
  if(plus==0){
    strncpy(tmp,&X[len1],len2+1);
  }
  else{
    strncpy(tmp,&X[len1+1],len2);
  }
  //printf("tmp:%s\n",tmp);
  tmp2 = atoi(tmp);
  //printf("tmp2:%d\n",tmp2);
  while(tmp2>0){
    decimal = decimal + (tmp2 %10)*base;
    tmp2 = tmp2/10;
    base = base*2;
  }
  
  //printf("%d\n",decimal);
  //return 0;
  
  //printf("%lu\n",strlen(decimalring));
  if(strlen(tmp)!=N){
    return 0;
  }
  
  // start decimal
  for(i=0;i<decimal+1;i++){
    k=i;
    //printf("i=%d,j=%d\n",i,j);
    //for(count=0;j!=i;count++)
      for(count=0;j!=i;count++){
        if(count!=0){
          k=j;
        }
        //printf("yes");
        if(k%2!=0){
          j=(k-1)/2;
          //      printf("i=%d,j=%d\n",i,j);
        }
        else{
          j=k/2+(int)pow(2,N-1);
          //printf("i=%d,j=%d\n",i,j);
        }
      }
      all_count += count;
      // printf("---%d is %d----\n",i,count);
      // printf("__________%d_________\n",all_count);
  }
  printf("%d\n",all_count%998244353);
    
  return 0;
}