Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

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

  ll N; cin >> N;
  string X; cin >> X;

  vector<ll> f(N + 10, 0);
  f[0] = 0;
  f[1] = 1;
  f[2] = 1;
  for (ll i = 3; i <= N; i++) {
    ll temp = i;
    ll p = 0;
    for (ll j = 0; j < N; j++) {
      p += (temp % 2);
      temp /= 2;
    }
    f[i] = f[(i % p)] + 1;
  }
  
  ll cnt = 0;
  for (ll i = 0; i < N; i++) {
    ll d = (X[i] - '0');
    cnt += d;
  }


  vector<vector<ll>> r_x(N + 10, vector<ll>(3, 0)); // 0 1 2
  vector<vector<ll>> r_2(N + 10, vector<ll>(3, 0));

  r_2[0][0] = 1;
  r_2[0][1] = 1;
  r_2[0][2] = 1;

  for (ll i = 1; i < N; i++) {
    r_2[i][1] = (r_2[i-1][1] * 2) % (cnt + 0);
    r_2[i][2] = (r_2[i-1][2] * 2) % (cnt + 1);
    if (cnt > 1) {
      r_2[i][0] = (r_2[i-1][0] * 2) % (cnt - 1);
    }
  }


  ll d = X[N - 1] - '0';

  if (cnt == 1) {
    r_x[0][0] = 0;
    r_x[0][1] = 0;
    r_x[0][2] = d;
  } else {
    r_x[0][0] = d % (cnt - 1);
    r_x[0][1] = d % (cnt + 0);
    r_x[0][2] = d % (cnt + 1);
  }
  for (ll i = 1; i < N; i++) {
    d = X[N - 1 - i] - '0';
    r_x[i][1] = (r_x[i-1][1] + d * r_2[i][1]) % (cnt + 0);
    r_x[i][2] = (r_x[i-1][2] + d * r_2[i][2]) % (cnt + 1);
    if (cnt > 1) {
      r_x[i][0] = (r_x[i-1][0] + d * r_2[i][0]) % (cnt - 1);
    }
  }

  for (ll i = 0; i < N; i++) {
    d = X[i] - '0';
    if (d == 1) {
      if (cnt == 1) {
        cout << 0 << '\n';
      } else {
        ll cnt_temp = cnt - 1;
        ll r = (r_x[N - 1][0] - r_2[N - 1 -i][0] + cnt_temp) % cnt_temp;
        cout << (f[r % cnt_temp] + 1) << '\n';
      }
    } else {
      ll cnt_temp = cnt + 1;
      ll r = (r_x[N - 1][2] + r_2[N - 1 -i][2] + cnt_temp) % cnt_temp;
      cout << (f[r % cnt_temp] + 1) << '\n';
    }
  }
  return 0;
}