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>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 5;
const int MOD = 998244353;

int n;
int f[MAXN],power[MAXN],s[MAXN];
vector<int> vec;
char str[MAXN];

int Calc(int len){
	int flag = 1,res = 0;
	for(int i = 1;i <= len;i++){
		if(s[i] == 1)
			res = (res + power[len - i]) % MOD;
	}
	for(int i = len + 1;i <= n;i++){
		if(s[i] != (s[i - len] ^ 1)){
			flag = (s[i] > (s[i - len] ^ 1) ? 1 : 0);
			break;
		}
	}
	return (flag + res) % MOD;
}

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	scanf("%d",&n);
	scanf("%s",str + 1);
	for(int i = 1;i <= n;i++)
		s[i] = str[i] - '0';
	power[0] = 1;
	for(int i = 1;i <= n;i++)
		power[i] = 2ll * power[i - 1] % MOD;
	for(int i = 1;i <= n;i++){
		if(n % i == 0 && (n / i) & 1)
			vec.push_back(i);
	} 
	for(int i = 0;i < (int)vec.size();i++)
		f[vec[i]] = Calc(vec[i]);
	for(int i = 0;i < (int)vec.size();i++){
		for(int j = 0;j < (int)vec.size();j++){
			if(vec[j] % vec[i] == 0 && vec[j] > vec[i])
				f[vec[j]] = (f[vec[j]] - f[vec[i]] + MOD) % MOD;
		}
	}
	int ans = 0;
	for(int i = 0;i < (int)vec.size();i++)
		ans = (ans + 2ll * vec[i] * f[vec[i]] % MOD) % MOD;
	printf("%d\n",ans);
	return 0;
}
