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;

constexpr long long MOD = 998244353;

int main() {
	int N;
	string S;
	cin >> N >> S;
	vector<int> P;
	for (int i = 1; i * i <= N; i++) {
		if (N % i == 0) {
			P.push_back(i);
			if (N / i != i) P.push_back(N / i);
		}
	}
	sort(P.begin(), P.end());
	vector<long long> ANS(N + 1);
	for (int L : P) {
		int X = N / L;
		if (X % 2 == 0) continue;
		string T = S.substr(0, L);
		long long COUNT = 0;
		for (int i = 0; i < L; i++) {
			COUNT *= 2;
			COUNT += T[i] - '0';
			COUNT %= MOD;
		}
		string U = "";
		for (int i = 0; i < N; i++) {
			if (i % (L * 2) < L) U += T[i % L];
			else U += T[i % L] ^ 1;
		}
		if (U <= S) COUNT++;
		COUNT %= MOD;
		ANS[L] = COUNT;
	}
	for (int L : P) {
		for (int K : P) {
			if (L > K&& L % K == 0) {
				ANS[L] += MOD - ANS[K];
				ANS[L] %= MOD;
			}
		}
	}
	long long SUM = 0;
	for (int L : P) {
		SUM += ANS[L] * L * 2;
		SUM %= MOD;
	}
	cout << SUM << endl;
}