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, n) for (int i = 0; i < (int)(n); ++i)
int calc(int d) {
  if (d == 0) {
    return 1;
  } else {
    return calc(d % bitset<32>(d).count()) + 1;
  }
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  vector<int> v1(n), v2(n);
  bitset<200000> b(s);
  int a = b.count(), a1 = a - 1, a2 = a + 1, b1 = 0, b2 = 0;
  v1[0] = 1, v2[0] = 1;
  if (b.test(0)) {
    b1 = 1, b2 = 1;
  }
  rep(i,n-1)
  {
    if (a > 1)
      v1[i + 1] = v1[i] * 2 % a1;
    v2[i + 1] = v2[i] * 2 % a2;
    if (b.test(i + 1)) {
      if (a > 1)
        b1 = (b1 + v1[i + 1]) % a1;
      b2 = (b2 + v2[i + 1]) % a2;
    }
  }
  for (int i = n - 1; i >= 0; i--)
    if (b.test(i))
      if (a > 1)
        cout << calc((b1 + a1 - v1[i]) % a1) << endl;
      else
        cout << 0 << endl;
    else
      cout << calc((b2 + a2 + v2[i]) % a2) << endl;
  return 0;
}