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<cstdio>
#include<vector>
using namespace std;
const int MAXN = 200000;
const int MOD = 998244353;
inline int add(int x, int y) {return (x + y)%MOD;}
inline int mul(int x, int y) {return 1LL*x*y%MOD;}
inline int sub(int x, int y) {return add(x,MOD-y);}
int pw2[MAXN + 5];
char X[MAXN + 5], Y[MAXN + 5];
int N;
int func(int x) {
	int ret = 1;
	for(int i=0;i<x;i++)
		if( X[i] - '0' ) ret = add(ret, pw2[x-i-1]);
	for(int i=0;i<N;i++)
		Y[i] = X[i % x]^((i/x)&1);
	for(int i=0;i<N;i++)
		if( X[i] < Y[i] ) {
			ret = sub(ret, 1);
			break;
		}
		else if( X[i] > Y[i] ) break;
	return ret;	
}
vector<int>v, v2;
int main() {
	int ans = 0; pw2[0] = 1;
	scanf("%d%s", &N, X);
	for(int i=1;i<=N;i++) pw2[i] = mul(2, pw2[i-1]);
	for(int i=1;i<=N;i++)
		if( N % i == 0 && (N / i) % 2 == 1 ) {
			int p = func(i);
			for(int j=0;j<v.size();j++)
				if( i % v2[j] == 0 ) p = sub(p, v[j]);
			v.push_back(p); v2.push_back(i);
			ans = add(ans, mul(p, mul(2, i)));
		}
	printf("%d\n", ans);
}