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 <algorithm>
#include <bitset>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>

using namespace std;

using ll = long long;
using ld = long double;

const ld pi = 3.141592653589793;

// int bitcount(int i) {
//   i = i - ((i >> 1) & 0x55555555);
//   i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
//   i = (i + (i >> 4)) & 0x0f0f0f0f;
//   i = i + (i >> 8);
//   i = i + (i >> 16);
//   return i & 0x3f;
// }

int f(int n) {
  int p = 0;
  bitset<32> bn(n);
  // cout << "@: " << bn << endl;
  while (bn.to_ulong() != 0) {
    bn = bitset<32>(bn.to_ulong() % bn.count());
    // cout << "!:" << bn << endl;
    p++;
  }
  return p;
}

int main() {
  int n, x;
  string xs;
  cin >> n >> xs;
  x = stoi(xs, NULL, 2);
  for (int i = 0; i < n; i++) {
    cout << f(x ^ (1 << (n - 1) >> i)) << endl;
  }
  // cout << bitset<16>(bitset<16>(8).to_ulong() % bitset<16>(8).count()) << endl;
  // cout << bitset<16>(8).count() << endl;
}