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;
const int BASE = 998244353;

int n;
string s;
map<int, int> m;

long long solve(int len)
{
  long long v = 0;
  for (int i = 0; i < len; i++)
    v = (v * 2 + s[i] - '0') % BASE;

  int isGood = 1;
  for (int i = len; i < n; i++)
  {
    int x = s[i % len] - '0';
    if (i / len % 2)
      x ^= 1;
    if (x == s[i] - '0')
      continue;
    if (x > s[i] - '0')
      isGood = 0;
    break;
  }

  v = (v + isGood) % BASE;
  long long res = 0;
  for (auto u : m)
    if (len % u.first == 0)
      v = (v - u.second + BASE) % BASE;

  m[len] = v;
  res = (res + 2LL * v * len) % BASE;
  return res;
}

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  cin >> n >> s;
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    if (n % i == 0 && n / i % 2)
      ans = (ans + solve(i)) % BASE;
  cout << ans << endl;
}
