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 ll = long long;
constexpr ll mod = 998244353;

int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  // dp[i] := i 回の操作で元に戻るものの個数
  vector<ll> dp(2 * n + 10);
  ll ret = 0;
  for (int k = 2; k <= 2 * n; k++) {
    if (2 * n % k) continue;
    if ((2 * n / k) % 2 == 0) continue;
    string t = s.substr(0, k / 2), t2 = t;
    cerr << k << " " << t << endl;
    bool f = true;
    for (int i = 0; i < k / 2; i++) {
      if (t[i] == '0')
        t2[i] = '1';
      else {
        t2[i] = '0';
        f = false;
      }
    }
    string x = "";
    while (x.size() < n) {
      x += t;
      x += t2;
    }
    x = x.substr(0, n);
    // cerr << k << " " << x << endl;
    if (x <= s) {
      dp[k]++;
    }
    if (!f) {
      ll sum = 0, tmp = 1;
      reverse(t.begin(), t.end());
      for (int i = 0; i < k / 2; i++) {
        if (t[i] == '1') sum = (sum + tmp) % mod;
        tmp = tmp * 2 % mod;
      }
      cerr << k << " " << sum << endl;
      dp[k] = (dp[k] + sum) % mod;
    }
    for (int j = 2; j * j <= k; j++) {
      if (k % j) continue;
      dp[k] = (dp[k] + mod - dp[j]) % mod;
      if (j * j != k) dp[k] = (dp[k] + mod - dp[k / j]) % mod;
    }
    ret += (ll)k * dp[k] % mod;
    ret %= mod;
    cerr << k << " " << dp[k] << endl;
  }
  cout << ret << endl;
}
