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;
#define all(V) V.begin(),V.end()
#define ll long long
#define MOD 998244353

ll fc[200010];

ll stoi2(string S) {
	ll a = 1, ans = 0;
	reverse(all(S));
	for (char c : S) {
		if (c == '1')ans += a;
		a *= 2;
		ans %= MOD;
		a %= MOD;
	}
	return ans;
}

int main(){
	int N;cin >> N;
	string X;cin >> X;
	ll a, b, c;
	string S, T, bs;
	fc[0] = 0;
	for (int i = 1;i <= N;i++) {
		fc[i] = 0;
		if (X[i - 1] == '0')S += '0', T += '1';
		else S += '1', T += '0';

		if (N%i)continue;
		a = N / i;
		if ((a & 1) == 0)continue;

		b = stoi2(S);bs = "";
		for (int j = 0;j < a;j++) {
			if (j & 1)bs += T;
			else bs += S;
		}

		if (X >= bs)
			b++;

		c = i;
		for (int j = 3;j <= c;j += 2) {
			if (c%j)continue;
			b -= fc[c / j];
			if (b < 0)b += MOD;
		}
		fc[i] = b;
	}

	ll ans = 0;
	for (int i = 1;i <= N;i++) {
		ans += fc[i] * i * 2;
		ans %= MOD;
	}
	cout << ans << endl;
}