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 be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=998244353, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

long long modpow(long long a, long long n) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}


int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    int N;
    string X;
    cin >> N >> X;
    vector<ll> f(N+1);
	for(int k=1;k<=N;k+=2)if(N%k==0){
		int L=N/k;
		bool fl=1;
		for(int i=0; i < N && fl ;i++){
			if(X[i]=='1'){
				if(i < L || X[i-L] == '1'){
					(f[L] += (i < L)? modpow(2LL,L-i-1): 1LL) %= mod;
				}
			}
			if(i >= L && X[i] == X[i-L]) fl = 0;
		}
		if(fl) (f[L] += 1) %= mod;
	}
	int ans=0;
	for(int i=1;i<=N;i++) if(N % i == 0 && N / i % 2 == 1 && f[i]){
		(ans += (f[i] * i * 2)%mod ) %= mod;
		for(int j=i*2;j<=N;j+=i) (f[j] -= f[i]) %= mod;
	}
    cout << (ans + mod) % mod << endl;
    return 0;
}
