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 , _ = 2e5 + 7;
char str[_] , tmp[_]; int num[_] , N;

signed main(){
	scanf("%d %s" , &N , str); int ans = 0; reverse(str , str + N);
	for(int i = 1 ; i <= N ; ++i)
		if(N % i == 0 && (N / i % 2)){
			int cnt = 0;
			for(int j = N - 1 ; j >= N - i ; --j) cnt = (cnt * 2 + str[j] - '0') % MOD;
			for(int j = N - 1 ; j >= 0 ; --j) tmp[j] = j >= N - i ? str[j] : (tmp[j + i] ^ 1);
			bool flg = 1;
			for(int j = N - 1 ; flg && j >= 0 ; --j){
				flg &= !(tmp[j] == '1' && str[j] == '0');
				if(str[j] == '1' && tmp[j] == '0') break;
			}
			num[i] = cnt + flg;
		}
	for(int i = 1 ; i <= N ; ++i)
		if(N % i == 0 && (N / i % 2)){
			ans = (ans + 1ll * num[i] * 2 * i) % MOD;
			for(int j = 2 ; j * i <= N ; ++j)
				num[j * i] = (num[j * i] - num[i] + MOD) % MOD;
		}
	cout << ans; return 0;
}
