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)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
#define INF (1e9)

vi A(2e5 + 10, -1);
ll N;

ll f(ll n) {
  if (n >= 2e5 + 10) {
    cout << "err " << n << endl;
    return -1;
  }
  if (A[n] >= 0) return A[n];
  ll cnt = __builtin_popcountll(n);
  ll next = n % cnt;
  A[n] = f(next) + 1;
  return A[n];
}

int main() {
  string X;
  cin >> N >> X;
  A[0] = 0;
  A[1] = 1;
  A[2] = 1;

  int cnt = 0;
  for (int i = 0; i < N; i++) {
    if (X[i] == '1') cnt++;
  }

  if (cnt == 0) {
    for (int i = 0; i < N; i++) {
      cout << 1 << endl;
    }
    return 0;
  }

  if (cnt == 1) {
    for (int i = 0; i < N; i++) {
      int a = (X[N - 1] == '1') ? 2 : 1;
      if (X[i] == '1')
        cout << 0 << endl;
      else
        cout << a << endl;
    }
    return 0;
  }

  // cnt > 1
  vi P(N), M(N);
  ll bitP = 1, bitM = 1, nextP = 0, nextM = 0;
  for (int i = N - 1; i >= 0; i--) {
    bitP %= cnt + 1, bitM %= cnt - 1;
    P[i] = bitP, M[i] = bitM;
    bitP *= 2, bitM *= 2;
    if (X[i] == '1') {
      nextP += P[i];
      nextP %= cnt + 1;
      nextM += M[i];
      nextM %= cnt - 1;
    }
  }

  vi ans(N);
  for (int i = 0; i < N; i++) {
    ll next;
    if (X[i] == '0') {
      next = nextP + P[i];
      next %= cnt + 1;
    } else {
      next = nextM - M[i];
      next += cnt - 1;
      next %= cnt - 1;
    }

    cout << f(next) + 1 << endl;
  }
}