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;
using Int = long long;
const Int MOD = 998244353;
array<Int, 1000010> P2;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    P2[0] = 1;
    for (int i = 1; i < P2.size(); i++) P2[i] = P2[i - 1] * 2 % MOD;
    int N; cin >> N;
    string X; cin >> X;
    string Y; Y.reserve(N);
    vector<int> A(N + 1);
    Int ans = 0;
    for (int n = 1; n <= N; n++) if (N % n == 0 && N % (2 * n) != 0) {
        Int cnt = 0;
        for (int i = 0; i < n; i++) {
            cnt = (cnt + P2[n - 1 - i] * (X[i] - '0')) % MOD;
        }
        Y.resize(2 * n);
        for (int i = 0; i < n; i++) Y[i] = X[i];
        for (int i = n; i < Y.size(); i++) Y[i] = '0' ^ '1' ^ X[i % n];
        bool ok = true;
        for (int i = 0; i < N; i++) {
            if (Y[i % Y.size()] == X[i]) continue;
            ok = Y[i % Y.size()] < X[i];
            break;
        }
        if (ok) cnt = (cnt + 1) % MOD;
        cnt = (cnt + MOD - A[n]) % MOD;
        for (int i = n + n; i < A.size(); i += n) {
            A[i] = (A[i] + cnt) % MOD;
        }
        ans = (ans + cnt * 2 * n) % MOD;
    }
    cout << ans << endl;
    return 0;
}
