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 LL;
const int N = 400000 + 10;
const LL MOD = 998244353;

int n;
char s[N], t[N];
LL pw[N], cnt[N];

LL cal(int x) {
	LL res = 0;
	for (int i = 1; i <= x / 2; i ++) {
		if (s[i] == '1')
			(res += pw[x/2 - i]) %= MOD;
	}
	for (int i = 1; i <= x / 2; i ++) t[i] = s[i];
	for (int i = x / 2 + 1; i <= 2 * n; i ++)
		t[i] = (char)((int)'0' + '1' - t[i - (x/2)]);
	//printf("%lld\n", res);
	//printf("%s\n", t + 1);
	bool ok = 1;
	for (int i = 1; i <= 2 * n; i ++)
		if (s[i] != t[i]) {
			if (t[i] > s[i]) {
				ok = false;
			}
			break;
		}
	if (ok) res ++;
	return res;
}
int main() {
	pw[0] = 1;
	for (int i = 1; i < N; i ++)
		pw[i] = pw[i-1] * 2 % MOD;

	scanf("%d%s", &n, s + 1);
	for (int i = 1; i <= n; i ++) {
		s[i + n] = (char)((int)'1' + '0' - s[i]);
	}
	for (int x = 1; x <= 2 * n; x ++) {
		if (2 * n % x == 0 && n % x) {
			cnt[x] = cal(x);
			//printf("cnt[%d] = %lld\n", x, cnt[x]);
		}
	}
	LL ans = 0;
	for (int i = 1; i <= 2 * n; i ++) {
		if (2 * n % i == 0 && n % i) {
			for (int j = i + i; j <= 2 * n; j += i) {
				cnt[j] -= cnt[i];
			}
			//printf("# cnt[%d] = %d\n", i, cnt[i]);
			cnt[i] = cnt[i] % MOD;
			ans += cnt[i] * i;
			ans = (ans % MOD + MOD) % MOD;
		}
	}
	cout << ans << endl;
}
