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 itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}

const int mod = 998244353;

int main() {
	int n = nxt();
	string s;
	cin >> s;

	vector<long long> deg(n + 1, 1);
	for (int i = 1; i <= n; ++i) {
		deg[i] = deg[i - 1] * 2 % mod;
	}
	vector<long long> cnt(n + 1);
	for (int d = 1; d <= n; d += 2) {
		if (n % d) {
			continue;
		}
		int l = n / d;
		long long tmp = 0;
		for (int i = 0; i < l; ++i) {
			if (s[i] == '1') {
				tmp += deg[l - 1 - i];
			}
		}
		string t = s;
		for (int i = l; i < n; ++i) {
			t[i] = t[i - l] == '0' ? '1' : '0';
		}
		if (t <= s) {
			tmp += 1;
		}
		tmp %= mod;
		cnt[d] = tmp;
		// ans += 2 * l * tmp % mod;
	}

	long long ans = 0;
	for (int i = n; i > 0; --i) {
		for (int j = i * 2; j <= n; j += i) {
			cnt[i] -= cnt[j];
		}
		cnt[i] %= mod;
		if (cnt[i] < 0) {
			cnt[i] += mod;
		}
		ans += (n / i) * 2 * cnt[i] % mod;
	}

	cout << ans % mod << "\n";

	return 0;
}
