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;
#define rep(i, a) for(int i=0; i<(a); i++)
typedef long long ll;

#ifdef _DEBUG
inline void dump() { cerr << endl; }
template<typename Head> void dump(Head&& head) { cerr << head; dump(); }
template<typename Head, typename... Tail> void dump(Head&& head, Tail&&... tail) { cerr << head << ", "; dump(forward<Tail>(tail)...); }
#define debug(...) do {cerr << __LINE__ << ":\t" << #__VA_ARGS__ << " = "; dump(__VA_ARGS__); } while (false)
#else
#define dump(...)
#define debug(...)
#endif

template<typename T>
struct edge {
  int src, to;
  T cost;
  edge(int to, T cost) : src(-1), to(to), cost(cost) {}
  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}
  edge &operator=(const int &x) {
    to = x;
    return *this;
  }
  operator int() const { return to; }
};
template<typename T>
using Edges = vector<edge<T>>;
template<typename T>
using WeightedGraph = vector<Edges<T>>;
using UnWeightedGraph = vector<vector<int>>;
template<typename T>
using Matrix = vector<vector<T>>;

/////////////////////////////////////////////////////////////////////

const ll inf = 1LL<<60;

ll f(ll n, ll p) {
  ll ret = 0;
  if (n == 0) return 0;
  ll t = n%p;
  ret += f(t, __builtin_popcountll(t));
  return ret + 1;
}

int main()
{
  ll n; cin >> n;
  string s; cin >> s;

  ll d = 0;
  for (int i=0; i<n; i++) {
    if (s[i] == '1') d++;
  }

  ll a = d-1;
  ll b = d+1;

  vector<ll> x(n, 1);
  vector<ll> y(n, 1);
  rep(i, n-1) {
    if (a > 0) x[n-i-2] = (x[n-i-1]*2) % a;
    if (b > 0) y[n-i-2] = (y[n-i-1]*2) % b;
  }

  ll xx = 0;
  ll yy = 0;
  rep(i, n) {
    if (s[i] == '1') {
      xx += x[i]; yy += y[i];
    }
  }

  for (int i=0; i<n; i++) {
    if (s[i] == '1') {
      if (a <= 0) {
        cout << 0 << endl;
        continue;
      }
      ll t = (xx + a - x[i]) % a;
      ll ans = f(t, __builtin_popcountll(t)) + 1;
      cout << ans << endl;
    } else {
      ll t = (yy + y[i]) % b;
      ll ans = f(t, __builtin_popcountll(t)) + 1;
      cout << ans << endl;
    }
  }

  return 0;
}