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;

typedef long long lint;

const int MAXN = 200010;
const int MOD = 998244353;

char X[MAXN];
int cnt[MAXN];
int two;

void addm(int &a, int b) {
	a += b;
	if(a >= MOD) a -= MOD;
}

void mnsm(int &a, int b) {
	a += MOD - b;
	if(a >= MOD) a -= MOD;
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int N;
	cin >> N >> X;
	
	int K, M;
	for(K = 0; N % (1 << K) == 0; K++);
	K--;
	M = N >> K;
	two = 1;
	for(int i = 1; i <= K + 1; i++) two = two * 2 % MOD;

	int ans = 0;
	vector<int> v;
	for(int i = 1; i * i <= M; i++) if(M % i == 0) {
		v.push_back(i);
		if(i * i != M) v.push_back(M / i);
	}
	sort(v.begin(), v.end());
	for(auto a : v) {
		for(int i = 0; i < (a << K); i++) cnt[a] = (cnt[a] * 2 + X[i] - '0') % MOD;
		bool b = false;
		for(int i = 0; i < N; i++) {
			int k = X[i % (a << K)] - '0';
			if(i / (a << K) % 2 == 1) k = 1 - k;
			if(k < X[i] - '0') break;
			if(k > X[i] - '0') {
				b = true;
				break;
			}
		}
		if(!b) cnt[a]++;
		for(auto aa : v) if(aa < a && a % aa == 0) mnsm(cnt[a], cnt[aa]);
		addm(ans, (lint)cnt[a] * a % MOD * two % MOD);
	}
	cout << ans << "\n";
	return 0;
}
