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)


#define INF 10000000000
#define MOD 1000000007

#if 1   // use_templates

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define rep(i, up) for (ll i = 0; i < up; ++i)
#define repp(i, up) for (ll i = 0; i <= up; ++i)
#define rrep(i, lo, up) for (ll i = lo; i < up; ++i)
#define rrepp(i, lo, up) for (ll i = lo; i <= up; ++i)
#define eep(i, lo, up) for ((i) = (lo); (i) < (up); ++(i))
#define eepp(i, lo, up) for ((i) = (lo); (i) <= (up); ++(i))


ll gcd(ll a, ll b);

ll gcd(ll a, ll b)
{
    if (a < b) {
        return gcd(b, a);
    }
    ll r;
    while (r = a % b) {
        a = b;
        b = r;
    }
    return b;
}

#endif  // end of use_templates

#include <bits/stdc++.h>

int main()
{
    ll n;
    cin >> n;
    string x;
    cin >> x;
    vector<ll> work(n);
    work[0] = 0;
    rrep(i, 1, n) {
        ll tmp = i % __builtin_popcount(i);
        work[i] = work[tmp] + 1;
    }
    ll pc = 0;
    rep(i, n) {
        if (x[i] == '1') {
            ++pc;
        }
    }

    if (pc == 1) {
        rep(i, n) {
            if (x[i] == '1') {
                cout << 0 << endl;
            } else if (i != n-1){
                cout << work[(x[n - 1] - '0') % 2] + 1 << endl;
            } else {
                cout << work[(x[n - 1] - '0' + 1) % 2] + 1 << endl;
            }
        }
        return 0;
    }

    ll pcp = pc + 1, pcn = pc - 1;
    ll mp = 0, mn = 0;
    rep(i, n) {
        mp = (mp * 2 + x[i] - '0') % pcp;
        mn = (mn * 2 + x[i] - '0') % pcn;
    }

    vector<ll> result(n);
    ll ap = 1, an = 1;
    for (ll i = n - 1; i >= 0; --i) {
        if (x[i] == '0') {
            result[i] = work[(mp + ap) % pcp] + 1;
        } else {
            result[i] = work[(mp - an + pcn) % pcn] + 1;
        }
        ap = ap * 2 % pcp;
        an = an * 2 % pcn;
    }
    rep(i, n) {
        cout << result[i] << endl;
    }
    return 0;
}