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;

ll N;
string X;
vector<ll> M;
vector<ll> W;
const ll MOD = 998244353;
ll R;

ll count(ll m) {
  ll x = 0;
  for (ll i = 0; i < m; ++i) {
    (x = 2 * x + (X[i] - '0')) %= MOD;
  }
  string y;
  for (ll i = 0; i < N; ++i) {
    char c = X[i % m];
    if ((i / m) % 2 == 1) c = c == '0' ? '1' : '0';
    y += c;
  }
  return y <= X ? (x + 1) % MOD : x;
}
int main() {
  cin >> N >> X;
  for (ll d = 1; d <= N; d += 2)
    if (N % d == 0) M.push_back(N / d);
  W.resize(M.size());
  for (ll i = 0; i < M.size(); ++i) {
    W[i] = 2 * M[i] % MOD;
    for (ll j = 0; j < i; ++j) {
      if (M[j] % M[i] == 0) (W[i] += MOD - W[j]) %= MOD;
    }
    (R += W[i] * count(M[i]) % MOD) %= MOD;
  }
  cout << R << endl;
  return 0;
}
