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;

map<int, int> memo;

int mod_pow(int x, int n, int mod) {
    int res = 1;
    while (n > 0) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void solve(int N, std::string X){
    int dp[N+1] = {};
    for (int i = 1; i <= N; i++) {
        int pc = __builtin_popcount(i);
        dp[i] = dp[i % pc] + 1;
        // cout << i << ":" << dp[i] << " ";
    }
    // cout << endl;

    int on = 0;
    for (int i = 0; i < N; i++) {
        if (X[i] == '1') {
            on++;
        }
    }
    // cout << "on: " << on << endl;
    int mod1 = on - 1;
    int mod2 = on + 1;

    int sum1 = 0;
    int sum2 = 0;
    for (int i = 0; i < N; i++) {
        if (X[i] == '1') {
            if (mod1) {
                sum1 = (sum1 + mod_pow(2, N - 1 - i, mod1)) % mod1;
            }
            sum2 = (sum2 + mod_pow(2, N - 1 - i, mod2)) % mod2;
        }
    }

    for (int i = 0; i < N; i++) {
        int first;
        if (X[i] == '1') {
            if (mod1) {
                first = (sum1 - mod_pow(2, N - 1 - i, mod1) + mod1) % mod1;
                // cout << first << "/" << mod1 << ":";
                cout << dp[first] + 1 << endl;
            } else {
                cout << 0 << endl;
            }
        } else {
            first = (sum2 + mod_pow(2, N - 1 - i, mod2) + mod2) % mod2;
                // cout << first << "/" << mod2 << ":";
            cout << dp[first] + 1 << endl;
        }
    }
}

// Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    int N;
    scanf("%d",&N);
    std::string X;
    std::cin >> X;
    solve(N, X);
    return 0;
}