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;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do { (t*=10)+=ch-'0'; ch=getchar(); } while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int mod=998244353;
const int maxn=(4e5)+10;
int n,s[maxn];
int T[maxn],d[maxn];
int f[maxn],ans;
char S[maxn];
void update(int &x,int y) {
	x+=y; if (x>=mod) x-=mod;
}
int solve(int k) {
	for (int i=1,x=1;i<=n;i++,x++) {
		if (x>2*k) x=1;
		if (x<=k) T[i]=d[x]; else T[i]=d[x-k]^1;
	}
	for (int i=1;i<=n;i++) {
		if (T[i]<d[i]) return s[k]+1;
		if (T[i]>d[i]) return s[k];
	}
	return s[k]+1;
}
int main() {
	//freopen("1.txt","r",stdin);
	read(n); scanf("%s",S+1);
	for (int k=1;k<=n;k++) {
		d[k]=S[k]-'0';
		s[k]=(s[k-1]*2+d[k])%mod;
	}
	for (int k=n;k>=1;k--)
		if (n%k==0&&(n/k)%2==1) {
			f[k]=2*k;
			for (int i=k*2;i<=n;i+=k) update(f[k],mod-f[i]);
			update(ans,(ll)f[k]*solve(k)%mod);
		}
	printf("%d\n",ans);
	return 0;
}