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>
using namespace std;

#define N 200002
const int mod = 998244353;

bool cmp(bitset<N> &y, bitset<N> &x) {
	for (int i = N - 1; i >= 0; i --) {
		if (y[i] != x[i]) return y[i] < x[i];
	}
	return true;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n;
	bitset<N> x;
	cin>>n>>x;
	vector<int> pre(n+1);
	for (int i = n - 1; i >= 0; i --) pre[i] = (2 * pre[i+1] + x[i]) % mod;
	vector<int> f(n+1);
	for (int d = 1; d <= n; d += 2) if (n % d == 0) {
		int l = n / d;
		f[d] = pre[n-l];
		bitset<N> y = x;
		for (int i = n - l - 1; i >= 0; i --) y[i] = !y[i+l];
		if (cmp(y, x)) {
			f[d] = (f[d] + 1) % mod;
		}
	}
	for (int i = ((n&1)?n:n-1); i >= 1; i -= 2) if (n % i == 0) {
		for (int j = 3 * i; j <= n; j += 2 * i) if (n % j == 0) {
			f[i] = (f[i] + mod - f[j]) % mod;
		}
	}
	int ans = 0;
	for (int i = 1; i <= n; i += 2) ans = (ans + 2ll * n / i * f[i]) % mod;
	cout<<ans;
}
