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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n;
char ch[200005], ns[400005];
int cans[400005], mu[400005];

int check(int x)
{
	if(n % x == 0) return 0;
	
	int ret = 0;
	rep(i, x / 2) {
		ret = 2LL * ret % MOD; 
		if(ch[i] == '1') ret = (ret + 1) % MOD;
	}
	
	rep(i, x) ns[i] = ch[i % (x / 2)] ^ (i / (x / 2));
	rep(i, n) ns[i] = ns[i % x];
	
	rep(i, n) if(ns[i] > ch[i]) return ret;
	else if(ns[i] < ch[i]) return (ret + 1) % MOD;
	return (ret + 1) % MOD;
}

int main()
{
	scanf("%d%s", &n, ch);
	int cur = 2 * n;
	
	for(int i = 1; i <= cur; i ++)
	if(cur % i == 0) cans[i] = check(i);
	
	mu[1] = 1;
	rep1(i, cur) for(int j = 2 * i; j <= cur; j += i)
	mu[j] -= mu[i];
	
	for(int i = cur; i >= 1; i --) for(int j = 2 * i; j <= cur; j += i)
	cans[j] = (cans[j] + 1LL * (MOD + mu[j / i]) * cans[i]) % MOD;
	
	int ans = 0;
	rep1(i, cur) if(cur % i == 0 && n % i != 0) ans = (ans + 1LL * cans[i] * i) % MOD;
	printf("%d\n", ans);
	return 0;
}