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<iostream>
#include<cstdio>
#include<algorithm>
#include<cassert>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD2 1000000007
#define MOD 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<int> divisors;

    int n;
    string str;

Int solve(int x){
    Int res = 0;
    for(int i = 0;i < x;i++){
        res *= 2;
        if(str[i] == '1')res++;
        res %= MOD;
    }
    bool exactly = true;
    for(int i = 0;i < n;i++){
        char c = str[i % x];
        if((i / x) % 2 == 1)c = '1' + '0' - c;
        if(str[i] == c)continue;
        if(str[i] < c)exactly = false;
        break;
    }
    if(exactly)res++;
    return res;
}

Int cnt[220000];
Int ans;
int main(){
    cin >> n;
    cin >> str;
    for(int i = 1;i <= n;i++){
        if(n % i != 0)continue;
        if((n / i) % 2 == 0)continue;               
        cnt[i] = solve(i);
        for(auto div: divisors){
            if(i % div != 0)continue;
            cnt[i] -= cnt[div];
            cnt[i] %= MOD;
        }
        ans += cnt[i] * i*2 % MOD;
        ans %= MOD;
        divisors.push_back(i);
    }
    if(ans < 0)ans += MOD;
    cout << ans << endl;
    return 0;
}