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: Runtime Error)

//#include <tourist>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> p;
const ll INF = 1e9;
const ll LINF = ll(1e18);
const ll MOD = 1000000007;
const ll dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const ll Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (ll i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v)          \
    cout << #v << ":";    \
    for (auto x : v)      \
    {                     \
        cout << x << ' '; \
    }                     \
    cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
//cout<<fixed<<setprecision(15);有効数字15桁
//-std=c++14
//-std=gnu++17
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll modp, modm;
string s;
ll mpow(ll x, ll n, ll M)
{ //x^n(mod M)
    ll ans = 1;
    while (n != 0)
    {
        if (n & 1)
            ans = ans * x % M;
        x = x * x % M;
        n = n >> 1;
    }
    return ans;
}
ll calc(ll m)
{
    ll res = 0;
    while (m)
    {
        ll popc = __builtin_popcount(m);
        m %= popc;
        res++;
    }
    return res;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n >> s;
    ll countt = 0;
    rep(i, n)
    {
        if (s[i] == '1')
            countt++;
    }
    ll count = countt;
    countt++;
    modp = 0;
    rep(i, n)
    {
        modp *= 2;
        if (s[i] == '1')
            modp++;
        modp %= countt;
    }
    countt -= 2;
    modm = 0;
    rep(i, n)
    {
        modm *= 2;
        if (s[i] == '1')
            modm++;
        modm %= countt;
    }
    countt++;
    rep(i, n)
    {
        if (s[i] == '0')
        {
            s[i] = '1';
            count++;
        }
        else
        {
            s[i] = '0';
            count--;
        }
        ll m;
        if (count > countt)
        {
            m = modp;
            m += mpow(2, s.size() - 1 - i, count);
            m%=count;
        }
        else
        {
            m = modm;
            m -= mpow(2, s.size() - 1 - i, count);
            m+=count;
            m%=count;
        }
        //cout<<m<<"\n";
        cout << calc(m) + 1 << "\n";
        if (s[i] == '0')
        {
            s[i] = '1';
            count++;
        }
        else
        {
            s[i] = '0';
            count--;
        }
    }
}