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 <vector>
#include <string>
using namespace std;
typedef long long ll;

const ll MOD = 998244353;

int main()
{
    ll n;
    string x;
    cin >> n >> x;
    vector<ll> v;
    ll d[200004] {0};
    for(ll i = n; i > 0; i--){
        if(i % 2 == 1 && n % i == 0) v.push_back(n / i);
    }
    ll ans = 0;
    for(ll l : v){
        ll sum = 0;
        ll b = 1;
        for(ll i = 0; i < l; i++){
            if(x[l - 1 - i] == '1') sum = (sum + b) % MOD;
            b = b * 2 % MOD;
        }
        string s = x.substr(0, l);
        string r;
        for(ll i = 0; i < l; i++){
            if(s[i] == '0') r += '1';
            else r += '0';
        }
        string coner = s;
        for(ll i = 0; i < (n / l) / 2; i++) coner += r + s;
        if(coner <= x) sum++;
        d[l] = (d[l] + sum) % MOD;
        for(ll y : v){
            if(y % l == 0 && y != l){
                d[y] = (d[y] - d[l] + MOD) % MOD;
            }
        }
        ans = (ans + d[l] * l * 2) % MOD;
    }
    cout << ans << endl;
}