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 <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 200005
#define ll long long
#define mod 998244353
char K[N];
int mul[N], f[N];
int n, ans, A[N];
void Getval(int x)
{
	for (int i = 1; i <= x; i++)
		if (A[i] == 1)
			f[x] = (f[x] + mul[x - i]) % mod;
	int flg = 0;
	for (int i = x + 1; i <= n; i++)
	{
		if (A[i] < !A[i - x])
		{
			flg = 1;
			break;
		}
		if (A[i] > !A[i - x])
			break;
	}
	if (!flg)
		f[x]++;
	for (int i = 1; i * i <= x; i++)
	{
		if (x % i == 0)
		{
			if (x != i && x / i % 2 == 1)
				f[x] = (f[x] - f[i]) % mod;
			if (i * i != x && i != 1 && i % 2 == 1)
				f[x] = (f[x] - f[x / i]) % mod;
		}
	}
	if (f[x] < 0)f[x] += mod;
}
int main()
{
	scanf("%d", &n);
	scanf("%s", K + 1);
	for (int i = 1; i <= n; i++)
		A[i] = K[i] - '0';
	mul[0] = 1;
	for (int i = 1; i <= n; i++)
		mul[i] = mul[i - 1] * 2ll % mod;
	for (int i = 1; i <= n; i++)
	{
		if (n % i == 0 && n / i % 2 == 1)
		{
			Getval(i);
			ans = (ans + f[i] * 2ll * i % mod) % mod;
		}
	}
	printf("%d\n", ans);
}
