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
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int mod=998244353,N=2e5+5;
int f[N],n,ans,a[N];
bool ok[N];

bool check(int x){
	for(int i=0,p=0;i<n;i+=x,p^=1) for(int j=1;j<=x;++j){
		if(a[i+j]>(a[j]^p)) return 1;
		if(a[i+j]<(a[j]^p)) return 0;
	}
	return 1;
}

void calc(int x){
	int tot=0;
	for(int i=1;i<=x;++i) tot=((tot<<1)+a[i])%mod;
	tot+=check(x);
	f[x]=(f[x]+tot)%mod;
	for(int i=x+x;i<=n;i+=x) if(ok[i]) f[i]=(f[i]-f[x]+mod)%mod;
}

signed main(){
	read(n);
	for(int i=1;i<=n;i++) scanf("%1d",&a[i]);
	for(int i=1;i<=n;i++) ok[i]=n%i==0&&(n/i&1);
	for(int i=1;i<=n;i++) if(ok[i]) calc(i);
	for(int i=1;i<=n;i++) ans=(ans+2ll*i*f[i])%mod;
	write(ans);
}