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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int MAXN = 200005;
const ll MOD = 998244353;

int n,tot;
int d[MAXN];

ll ans;
ll cnt[MAXN];

string s;

int main()
{
	cin >> n >> s;
	int pw2 = 1,N = n;
	while (~N & 1)
	{
		pw2 <<= 1;
		N >>= 1;
	}
	for (int i = 1;i <= N;i++)
		if (N % i == 0)
		{
			d[++tot] = pw2 * i;
			for (int j = 0;j < d[tot];j++)
				cnt[tot] = (cnt[tot] * 2 + s[j] - '0') % MOD;
			cnt[tot]++;
			for (int j = 0;j < s.size();j++)
				if (s[j] != ((s[j % d[tot]] - '0') ^ ((j / d[tot]) & 1)) + '0')
				{
					if (s[j] < ((s[j % d[tot]] - '0') ^ ((j / d[tot]) & 1)) + '0')
						cnt[tot]--;
					break;
				}
		}
	for (int i = 1;i <= tot;i++)
		for (int j = i - 1;j >= 1;j--)
			if (d[i] % d[j] == 0)
				(cnt[i] -= cnt[j]) %= MOD;
	for (int i = 1;i <= tot;i++)
		(ans += d[i] * 2 * cnt[i]) %= MOD;
	printf("%lld\n",(ans + MOD) % MOD);
	return 0;
}