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 ll long long 
using namespace std;

const int maxn=2e5+10;
const int mod=998244353;
char c[maxn];
ll a[maxn],sum[maxn],tot,f[maxn],b[maxn];
int n;
ll poww(ll x,ll k) {
	ll ans=1;
	while(k) {
		if(k&1) ans=ans*x%mod;x=x*x%mod;k>>=1; 
	}return ans;
}
bool check(int x) {
	int now=1,f=0;
	for(int i=1;i<=n;++i) {
		if(((c[now]-'0')^f)>(c[i]-'0')) return 0;
		else if(((c[now]-'0')^f)<(c[i]-'0')) return 1;
		else {
			now++;
			if(now>x) now=1,f^=1;
		}
	}
	return 1;
}
int get(int x) { 
	if((n/x)%2) {
		if(check(x))
		return sum[x]+1;
		else return sum[x];
	}
	return 0;
} 
ll ans;
int main()
{
	scanf("%d",&n);
	scanf("%s",c+1);
	for(int i=1;i<=n;++i) {
		tot=tot*2%mod;
		if(c[i]=='1') tot=tot+1;
		sum[i]=tot; 
	//	cout<<sum[i]<<"*\n";
	}
	tot=tot%mod;
	for(int i=1;i<=sqrt(n);++i) {
		if(n%i==0) f[i]=get(i);
		if(i*i!=n&&(n%(n/i)==0)) f[n/i]=get(n/i);
	}
	//cout<<get(n/2)<<"\n";
	//for(int i=1;i<=n;++i) cout<<f[i]<<"*\n";
	for(int i=1;i<=n;++i) {
		for(int j=i+i;j<=n;j+=i) f[j]=(f[j]-f[i]+mod)%mod;
	}
	for(int i=1;i<=n;++i) {
		if(n%i==0) {
			ans=(ans+f[i]*2*i%mod)%mod;
		}
	}
	cout<<ans<<"\n";
	return 0;
}
/*
5
00100
*/