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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int MAX = 200001;

int main() {
  int n;
  string x;
  cin >> n >> x;

  int pop_count_origin = count(x.begin(), x.end(), '1');
  int pop_count_m = pop_count_origin - 1;
  int pop_count_p = pop_count_origin + 1;

  int x_m_origin = 0;
  int x_p_origin = 0;
  rep(i, n) {
    x_m_origin = (x_m_origin * 2 + int(x[i] - '0')) % pop_count_m;
    x_p_origin = (x_p_origin * 2 + int(x[i] - '0')) % pop_count_p;
  }

  vector<int> pow_m(MAX, 1);
  vector<int> pow_p(MAX, 1);
  for (int i = 1; i < MAX; i++) {
    pow_m[i] = (pow_m[i - 1] * 2) % pop_count_m;
    pow_p[i] = (pow_p[i - 1] * 2) % pop_count_p;
  }

  vector<int> f(MAX);
  vector<int> pop_count(MAX);
  for (int i = 1; i < MAX; i++) {
    pop_count[i] = pop_count[i / 2] + i % 2;
    f[i] = f[i % pop_count[i]] + 1;
  }

  rep(i, n) {
    if (x[i] == '1') {
      int next = (x_m_origin - pow_m[n - 1 - i]) % pop_count_m;
      if (next < 0) next += pop_count_m;
      cout << f[next] + 1 << endl;
    } else {  // x[i] == 1
      int next = (x_p_origin - pow_p[n - 1 - i]) % pop_count_p;
      if (next < 0) next += pop_count_p;
      cout << f[next] + 1 << endl;
    }
  }

  return 0;
}