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>
using namespace std;
using ll = long long;
using ld = long double;
namespace my {
char itoc(int i) { return i + '0'; }
int ctoi(char c) { return c - '0'; }
ll gcd(ll a, ll b) {
  a = abs(a);
  b = abs(b);
  if (a < b) swap(a, b);
  while (b != 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
ll lcm(ll a, ll b) { return a * b / my::gcd(a, b); }
vector<pair<ll, ll>> primeFac(ll n) {
  vector<pair<ll, ll>> ret;
  for (ll i = 2; i * i <= n; i++) {
    if (n % i != 0) continue;
    ll exp = 0;
    while (n % i == 0) {
      n /= i;
      exp++;
    }
    ret.push_back(make_pair(i, exp));
  }
  if (n != 1) ret.push_back(make_pair(n, 1));
  return ret;
}
}  // namespace my
   ////////////////////////

const int N = 2e5;

int f(bitset<N> bs) {
  int cnt = 0;
  while (bs.to_ullong() != 0) {
    int ppc = bs.count();
    int tmp = bs.to_ullong() % ppc;
    bitset<N> bs_(tmp);
    bs = bs_;
    cnt++;
  }
  // cout << cnt << endl;
  return cnt;
}

int main() {
  int n;
  cin >> n;
  bitset<N> tmp;
  cin >> tmp;
  bitset<N> bs = tmp;
  vector<int> ans(n);

  // cout << bs << endl;

  for (int i = 0; i < n; i++) {
    // cout << n - 1 - i << endl;
    bs.flip(n - 1 - i);
    ans.at(i) = f(bs);
    bs.flip(n - 1 - i);
  }

  for (int i = 0; i < n; i++) {
    cout << ans.at(i) << endl;
  }
}