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: Wrong Answer)

#pragma GCC target("popcnt")
#include <x86intrin.h>

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

#ifdef LOCAL
#include "debug.h"
#else
#define DEBUG(...)
#endif

int f(int n) {
  if (n == 0) {
    return 1;
  }
  return f(n % _popcnt64(n)) + 1;
}

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  string s;
  cin >> s;
  int c = count(begin(s), end(s), '1');
  vector<int> mods{c + 1, c - 1};
  vector a(2, vector(n, 1));
  for (int t : {0, 1}) {
    for (int i = n - 1; i--; ) {
      a[t][i] = 2 * a[t][i + 1];
      if (a[t][i] >= mods[t]) {
        a[t][i] -= mods[t];
      }
    }
  }
  vector<int> base(2);
  for (int t : {0, 1}) {
    for (int i = 0; i < n; ++i) {
      if (s[i] == '0') continue;
      base[t] += a[t][i];
      if (base[t] >= mods[t]) {
        base[t] -= mods[t];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      int cur = base[0] + a[0][i];
      if (cur >= mods[0]) {
        cur -= mods[0];
      }
      cout << f(cur) << '\n';
    } else {
      int cur = base[1] - a[1][i];
      if (cur < 0) {
        cur += mods[1];
      }
      cout << f(cur) << '\n';
    }
  }
}