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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

ll mod = 998244353;

int n, x[300000];
ll r, s[300000], p2[300000];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  char c;
  p2[0] = 1;
  for (int i = 0; i < n; i++)
    cin >> c, x[i] = c == '1', p2[i + 1] = (p2[i] * 2) % mod;
  for (int d = 1; d <= n; d++) {
    if ((n % d) == 0 && (n % (2 * d)) != 0) {
      ll k = 2 * d;
      ll h = 0;
      for (int j = -1; j < d - 1; j++)
        if (x[j + 1] == 1)
          h += p2[d - j - 2];
      bool sm = true;
      for (int j = d; j < n; j++)
        if ((x[j % d] ^ ((j / d) & 1)) != x[j]) {
          sm = (x[j % d] ^ ((j / d) & 1)) < x[j];
          break;
        }
      if (sm)
        h++;
      h %= mod;
      h = (h + mod - s[d]) % mod;
      r = (r + h * k) % mod;
      for (int j = 2; d * j <= n; j++)
        s[d * j] = (s[d * j] + h) % mod;
    }
  }
  cout << r << "\n";
}
