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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define mod 998244353

using namespace std;
typedef long long ll;
const int MAXN = 2000005;
int n,a[MAXN],b[MAXN];
char str[MAXN];
ll f[MAXN];
inline void solve(int x)
{
	ll res = 0;
	for(int i = 1; i<=x/2; i++)
		res = (res*2+a[i])%mod;
	res = (res+1)%mod;
	for(int i = 1, j = 0; i<=n; i += x/2, j++)
		for(int k = i; k<=i+x/2-1; k++)
			b[k] = a[k-i+1]^(j&1);
	bool flag = true;
	for(int i = 1; i<=n; i++)
	{
		if(b[i]>a[i])
			flag = false;
	    else if(b[i]<a[i])
	    	break;
	}
	if(!flag)
		res = (res+mod-1)%mod;
	f[x] = res;
}

int main()
{
	cin >> n >> str+1;
	for(int i = 1; i<=n; i++)
		a[i] = str[i]-'0';
	for(int i = 1; i<=2*n; i++)
		if(2*n%i==0&&n%i!=0)
			solve(i);
	for(int i = 1; i<=2*n; i++)
		if(2*n%i==0&&n%i!=0)
			for(int j = 2*i; j<=2*n; j += i)
				if(2*n%j==0&&n%j!=0)
					f[j] = (f[j]-f[i]+mod)%mod;
	ll ans = 0;
	for(int i = 1; i<=2*n; i++)
		ans = (ans+f[i]*i)%mod;
	cout << ans << endl;
	return 0;
}
