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;
typedef long long LL;

const int MAXN = 1e6+7;
int dp[MAXN];
//LL prefix[MAXN], suffix[MAXN];
int f[MAXN];

int N, now;
string X;
void solve(char c) {
    int mod;
    if (c=='1') {
        if (now==0) return;
        mod = now-1;
    } else {
        if (now==N) return;
        mod = now+1;
    }

    if (mod == 0) {
        assert(c=='1');
        for (int i = 1; i <= N; i++) {
            if (X[i-1]=='1') {
                f[i] = 0;
            }
        }
        return;
    }

//    cout << c << " mod " << mod << endl;

    int total = 0;
    for (int i = N, pw = 1; i > 0; i--, pw = (pw+pw)%mod) {
        if (X[i-1]=='1') total = (total+pw)%mod;
    }

    for (int i = N, pw = 1; i > 0; i--, pw = (pw+pw)%mod) {
        if (X[i-1]!=c) continue;
        int tmp = total;
        if (c=='1') tmp = (tmp+mod-pw)%mod;
        else tmp = (tmp+pw)%mod;
        f[i] = dp[tmp]+1;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    for (int i = 1; i < MAXN; i++) {
        dp[i] = dp[i%__builtin_popcount(i)]+1;
    }

    cin >> N >> X;
    now = 0;
    for (char c : X) now += c=='1';

    solve('0');
    solve('1');

    for (int i = 1; i <= N; i++) cout << f[i] << "\n";



    return 0;
}