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;
const int mod = 998244353;
const int maxn = 200200;
int mul(long long a, long long b){
	return a * b % mod;
}
int p2[maxn];
int main(){
	p2[0] = 1;
	for(int i = 1; i < maxn; i++) p2[i] = mul(2, p2[i - 1]);
	int n; cin >> n;
	string s; cin >> s;
	vector<int> divs;
	for(int d = 1; d <= n; d++){
		if(n % d == 0 && (n / d) % 2 == 1){
			divs.push_back(d);
		}
	}
	int len = divs.size();
	vector<int> how(len);
	int ans = 0;
	for(int i = 0; i < len; i++){
		int d = divs[i];
		int ways = 0;
		for(int j = 0; j < d; j++){
			if(s[j] == '1'){
				ways += p2[d - j - 1];
				if(ways >= mod) ways -= mod;
			}
		}
		bool ok = true;
		for(int j = d; j < n && ok; j++){
			int dist = j / d;
			if(dist % 2){
				if(s[j % d] == '0'){
					if(s[j] == '0'){
						ok = false;
					}
				} else {
					if(s[j] == '1'){
						break;
					}
				}
			} else {
				if(s[j % d] == '0'){
					if(s[j] == '1'){
						break;
					}
				} else {
					if(s[j] == '0'){
						ok = false;
					}
				}
			}
		}
		if(ok){
			ways += 1;
			if(ways >= mod) ways -= mod;
		}
		for(int j = 0; j < i; j++){
			if(d % divs[j] == 0){
				ways += mod - how[j];
				if(ways >= mod) ways -= mod;
			}
		}
		how[i] = ways;
		ans += mul(ways, d);
		if(ans >= mod) ans -= mod;
	}
	cout << mul(ans, 2) << endl;
	return 0;
}
