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 rep(i,n) for (int i=0; i<(int)(n); ++i)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using Graph = vector<vi>;  // vi <=> vl
using P = pair<int, int>;  // int <=> ll
const int MOD = 998244353;
const double EPS = 1e-9;

int main() {
  int n; cin >> n;
  bitset<200000> s; cin >> s;

  vl power(n);
  power[0] = 1;
  rep(i,n-1) power[i+1] = power[i] * 2 % MOD;
  ll num = 1;
  rep(i,n) if (s.test(i)) (num += power[i]) %= MOD;
  ll ans = n*2 * num % MOD;

  map<int, ll> decrease;
  ll tmp = 0;

  for (int i=1; i<=n/3; ++i) {
    ((tmp *= 2) += s[n-i]) %= MOD;

    if (n%i == 0 && n/i%2 == 1) {
      decrease[i] = tmp;

      bool x = true;
      bool exit = false;

      for (int j=1; j<n/i; ++j) {
        for (int k=1; k<=i; ++k) {
          int bit = s[n-k];
          if (j%2) bit = 1 - bit;

          if (s[n-j*i-k] == bit) continue;

          if (s[n-j*i-k] < bit) x = false;
          exit = true;
          break;
        }
        if (exit) break;
      }

      if (x) ++decrease[i];
    }
  }

  vi length;

  for (auto p : decrease) {
    int key = p.first;
    
    for (auto d : length) {
      if (key%d == 0 && key/d%2 == 1) {
        decrease[key] -= decrease[d];
        if (decrease[key] < 0) decrease[key] += MOD;
      }
    }

    ans -= (n*2 - key*2) * decrease[key] % MOD;
    if (ans < 0) ans += MOD;
    length.push_back(key);
  }

  cout << ans << endl;
  return 0;
}
