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;

ll mod;
struct mint {
  ll x;  // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

ll modpow(ll a, ll n, ll m) {
  ll result = 1;

  while (n > 0) {
    if (n & 1) result = result * a % m;
    a = a * a % m;
    n >>= 1;
  }
  return result;
}

int f(int n, int count) {
  if (n == 0) return count;
  return f(n % __builtin_popcount(n), count + 1);
}

mint getrem(string X, ll icount) {
  mint rem = 0;
  for (int i = 0; i < X.size(); i++) {
    if (X[i] == '1') {
      rem += modpow(2, X.size() - i - 1, icount);
    }
  }
  return rem;
}

int main() {
  string X;
  int N;
  cin >> N >> X;

  ll icount = 0;

  rep(i, N) {
    if (X.at(i) == '1') icount++;
  }

  mod = icount - 1;
  if (mod == 0) mod = 1;  // foo
  mint lessrem = mod > 0 ? getrem(X, mod) : 0;
  mod = icount + 1;
  mint morerem = getrem(X, icount + 1);

  rep(i, N) {
    mod = 1;  // foo
    mint rem;
    if (X.at(i) == '1') {
      rem = lessrem;
      mod = icount - 1;
      if (mod == 0) {
        cout << 0 << endl;
        continue;
      }
      rem -= modpow(2, X.size() - i - 1, mod);
    } else {
      rem = morerem;
      mod = icount + 1;
      rem += modpow(2, X.size() - i - 1, mod);
    }

    cout << f(rem.x, 0) + 1 << endl;
  }

  return 0;
}