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>
#define MOD 998244353

using namespace std;

typedef long long ll;

char s1[200005],s2[200005];
int f[200005];

int main() {
  int n;
  scanf("%d%s",&n,s1);
  reverse(s1,s1+n);
  for(int i=1;i<=n;i++)
    if (n%i==0&&((n/i)&1)) {
    	int s=0;
    	for(int j=n-1;j>=n-i;j--) s=(s*2LL+s1[j]-'0')%MOD;
    	f[i]=s;
    	for(int j=n-1;j>=0;j--)
    		if (j>=n-i) s2[j]=s1[j]; else s2[j]=((s2[j+i]-'0')^1)+'0';
    	bool ok=1;
    	for(int j=n-1;j>=0;j--)
    	  if (s1[j]!=s2[j]) {
    	  	if (s2[j]>s1[j]) ok=0;
    	  	break;
		  }
		f[i]=(f[i]+ok)%MOD;
	}
  int ans=0;
  for(int i=1;i<=n;i++)
    if (n%i==0) {
    	for(int j=2;i*j<=n;j++) f[i*j]=(f[i*j]-f[i]+MOD)%MOD;
    	ans=(ans+f[i]*2LL%MOD*i)%MOD;
	}
  printf("%d\n",ans);
  return 0;
}