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;
typedef long long ll;
const int MAXN = 200100;
const ll MOD = 998244353;

int N;
string X;
vector <int> pfac;
ll rv[MAXN];

ll gogo (int K)
{
    ll cv = 0;
    for (int i = 0; i < K; i++)
        cv = (2 * cv + (X[i] - '0')) % MOD;
    if (K == N) cv++;
    for (int i = K; i < N; i++)
    {
        int ov = 1 - (X[i-K] - '0');
        int nv = X[i] - '0';
        if (ov > nv)
        {
            break;
        }
        else if (ov < nv)
        {
            cv++;
            break;
        }
        else if (i == N - 1)
        {
            cv++;
            break;
        }
    }
    cv %= MOD;
    return cv;
}

int main()
{
    ios_base::sync_with_stdio(0);

    cin >> N;
    cin >> X;

    for (int i = 1; i <= N; i++)
    {
        if (N % i == 0)
        {
            if ((N / i) % 2 == 1)
                pfac.push_back(i);
        }
    }

    for (int i = 0; i < pfac.size(); i++)
    {
        rv[i] = gogo (pfac[i]);
        //cout << pfac[i] << " " << rv[i] << "\n";
    }

    for (int i = 0; i < pfac.size(); i++)
    {
        for (int j = i + 1; j < pfac.size(); j++)
            if (pfac[j] % pfac[i] == 0)
            {
                rv[j] -= rv[i];
            }
    }

    ll ans = 0;
    for (int i = 0; i < pfac.size(); i++)
    {
        //cout << pfac[i] << " " << rv[i] << "\n";
        ans = ans + (2 * pfac[i] * rv[i]);
        ans %= MOD;
    }
    ans = (ans + MOD) % MOD;
    cout << ans << "\n";
}