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: Accepted)

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

int main() {
  int N;
  string X;
  cin >> N >> X;
  int pcnt = 0;
  for (int i = 0; i < N; i++) {
    if (X[i] == '1') pcnt++;
  }
  ll X_minus = 0, X_plus = 0;
  vector<ll> p2_minus(N, 1), p2_plus(N, 1);
  for (int i = N-1; i >= 0; i--) {
    X_minus += (X[i] - '0') * p2_minus[i];
    X_plus += (X[i] - '0') * p2_plus[i];
    if (i > 0) {
      p2_minus[i-1] = p2_minus[i]*2;
      if (pcnt != 1) p2_minus[i-1] %= pcnt-1;
      p2_plus[i-1] = p2_plus[i]*2;
      p2_plus[i-1] %= pcnt+1;
    }
  }
  vector<int> ans(N, 0);
  for (int i = 0; i < N; i++) {
    if (X[i] - '0' == 1 && pcnt == 1) {
      cout << 0 << endl;
      continue;
    }
    int pcnt_re;
    ll X_tmp;
    if (X[i] == '0') {
      X[i] = '1';
      pcnt_re = pcnt+1;
      X_tmp = X_plus + p2_plus[i];
      X_tmp %= pcnt_re;
    }
    else {
      X[i] = '0';
      pcnt_re = pcnt-1;
      X_tmp = X_minus - p2_minus[i] + pcnt_re;
      if (pcnt_re != 0) X_tmp %= pcnt_re;
    }
    ans[i]++;
    while (X_tmp > 0) {
      X_tmp = X_tmp % __builtin_popcount(X_tmp);
      ans[i]++;
    }
    cout << ans[i] << endl;
  }
}