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;

vector<int64_t> memo;
int64_t f(int64_t n) {
    if (n == 0) {
        return 0;
    }
    if (memo[n] != -1) {
        return memo[n];
    }

    return memo[n] = (f(n % __builtin_popcountll(n)) + 1);
}

//a^b % modを求める
int64_t MODpow(int64_t a, int64_t b, int64_t mod) {
    int64_t result = 1;
    while (b) {
        if (b % 2 == 1) {
            result *= a;
            result %= mod;
        }

        b /= 2;
        a *= a;
        a %= mod;
    }

    return result;
}

int main() {
    int64_t N;
    cin >> N;

    memo.resize(N + 2, -1);

    string X;
    cin >> X;

    int64_t b = count(X.begin(), X.end(), '1');
    //ビット数はb or (b + 1)になる

    if (b == 1) {
        //b - 1が0になって終わるので適宜計算
        int64_t one_index = find(X.begin(), X.end(), '1') - X.begin();

        for (int64_t i = 0; i < N; i++) {
            if (i == one_index) {
                //0になるので0
                cout << 0 << endl;
            } else {
                //こことone_indexだけが立つ
                int64_t v = (MODpow(2, N - 1 - i, 2) + MODpow(2, N - 1 - one_index, 2)) % 2;
                cout << f(v) + 1 << endl;
            }
        }

        return 0;
    }

    int64_t case_b_minus_1 = 0, case_b_plus_1 = 0;
    for (int64_t i = 0; i < N; i++) {
        if (X[i] == '1') {
            (case_b_minus_1 += MODpow(2, N - 1 - i, b - 1)) %= (b - 1);
            (case_b_plus_1 += MODpow(2, N - 1 - i, b + 1)) %= (b + 1);
        }
    }

    for (int64_t i = 0; i < N; i++) {
        //iを反転
        if (X[i] == '0') {
            //1になる
            int64_t v = (case_b_plus_1 + MODpow(2, N - 1 - i, b + 1)) % (b + 1);
            cout << f(v) + 1 << endl;
        } else {
            //0になる
            int64_t v = case_b_minus_1;
            //自分の影響を除く
            (v += b - 1 - MODpow(2, N - 1 - i, b - 1)) %= (b - 1);

            cout << f(v) + 1 << endl;
        }
    }
}