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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int f(int n) {
    int k = 0;
    while (n != 0) {
        n %= bitset<32>(n).count();
        k++;
    }
    return k;
}
int pow_mod(int a, int b, int mod) {
    if (b == 0) return 1 % mod;
    int k = pow_mod(a, b/2, mod);
    k *= k;
    k %= mod;
    if (b % 2 == 0) return k;
    return (k*a) % mod;
}
int main() {
    int N;
    string X;
    cin >> N >> X;
    reverse(X.begin(), X.end());
    int s = 0;
    for (int i = 0; i < N; i++) if (X[i] == '1') s++;
    int A = 0, B = 0;
    for (int i = 0; i < N; i++) {
        if (X[i] == '1') {
            A += pow_mod(2, i, s+1);
            A %= s+1;
            if (s-1 > 0) {
                B += pow_mod(2, i, s-1);
                B %= s-1;
            }
        }
    }
    for (int i = N-1; i >= 0; i--) {
        if (X[i] == '1') {
            if (s-1 == 0) {
                cout << 0 << endl;
                continue;
            }
            int b = B;
            b -= pow_mod(2, i, s-1);
            b += s-1;
            b %= s-1;
            cout << 1+f(b) << endl;
        }
        else {
            int a = A;
            a += pow_mod(2, i, s+1);
            a %= s+1;
            cout << 1+f(a) << endl;
        }
    }
}