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 <bits/stdc++.h>
const int INF = 1e9;
const int MOD = 1e9+7;
const long long LINF = 1e18;
#define dump(x)  cout << 'x' << ' = ' << (x) << ` `;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
typedef long long ll;
using namespace std;
typedef pair<ll, ll> P;

template<typename T>
void print(const vector<T> &x) {
    int n = x.size();
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x) {
    int n = x.size();
    rep(i,n) {
        rep(j,x[i].size()) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

template<typename T>
void print(const vector<T> &x, int n) {
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x, int n, int m) {
    rep(i,n) {
        rep(j,m) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

void input_init() {
    cin.tie(0); ios::sync_with_stdio(false);
}

// x^n(mod M)
ll mpow(ll x, ll n, ll mod) {
    ll ans = 1ll;
    while(n != 0){
        if (n&1) ans = ans*x%mod;
        x = x*x%mod;
        n = n >> 1;
    }
    return ans;
}

template< int mod >
struct ModInt {
    int x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod) x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod) x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p) {
        x = (int) (1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
        t = a / b;
        swap(a -= t * b, b);
        swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
        if(n & 1) ret *= mul;
        mul *= mul;
        n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt< mod >(t);
        return (is);
    }

    static int get_mod() { return mod; }
};

using modint = ModInt<MOD>;

int main(int argc, char const *argv[]) {
    int n; cin>>n;
    string s; cin>>s;
    vector<ll> a(n+1);
    auto calc = [&](int x) {
        int ret = 0;
        while (x > 0) {
            int pop = __builtin_popcount(x);
            x = x%pop;
            ret++;
        }
        return ret;
    };
    for (int i = 0; i <= n; ++i) a[i] = calc(i);

    int cnt = 0;
    rep(i,n) if (s[i]=='1') cnt++;
    int mod_1 = cnt+1; int mod_2 = cnt-1;
    ll sum_1 = 0; ll sum_2 = 0;
    reverse(s.begin(), s.end());

    rep(i,n) {
        if (s[i] == '0') continue;
        sum_1 += mpow(2,i,mod_1)%mod_1;
        sum_1 %= mod_1;
        sum_2 += mpow(2,i,mod_2)%mod_2;
        sum_2 %= mod_2;
    }

    vector<ll> ans(n);
    rep(i,n) {
        if (s[i]=='0') {
            ll tmp = sum_1;
            tmp += mpow(2,i,mod_1);
            tmp %= mod_1;
            ans[i] = a[tmp]+1;
        }
        else {
            ll tmp = sum_2;
            tmp -= mpow(2,i,mod_2);
            tmp %= mod_2;
            if (tmp < 0) tmp += mod_2;
            ans[i] = a[tmp]+1;
        }
    }
    reverse(ans.begin(), ans.end());
    rep(i,n) cout << ans[i] << '\n';
    return 0;
}