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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
 
#define $(x) (int)((x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define endl '\n'
#define x first
#define y second
#define int long long
#define double long double
#define pii pair<int, int>
#define pb push_back
#define vec vector
#define beg begin
#define dbg(x) cout << #x << " = " << x << endl;
 
template<class T> ostream& operator<<(ostream &str, vector<T> &a) {
    for (auto &i : a) {
        str << i << " ";
    }
    return str;
}
 
template<class T> istream& operator>>(istream &str, vector<T> &a) {
    for (auto &i : a) {
        str >> i;
    }
    return str;
}
 
template<class T> ostream& operator<<(ostream &str, pair<T, T> &a) {
    str << a.first << " " << a.second;
    return str;
}
 
template<class T> istream& operator>>(istream &str, pair<T, T> &a) {
    str >> a.first >> a.second;
    return str;
}
 
void solve(); signed main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    srand(time(0));
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cout.setf(ios::fixed); cout.precision(6);
    solve();    
    return 0;   
}
 
const int MOD = 1e9 + 7, MAXN = 5e5 + 10, INF = 2e9 + 10, BASE = 37; 

int bin_pow(int num, int n, int mod) {
    if (n == 0) {
        return 1;
    }
    if (n & 1) {
        return num * bin_pow(num, n - 1, mod) % mod;
    }
    else {
        int tmp = bin_pow(num, n / 2, mod);
        return tmp * tmp % mod;
    }
}

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
        if (s[i] == '1') {
            cnt++;
        }
    }
    int mod1 = cnt - 1, mod2 = cnt, mod3 = cnt + 1;

    assert(mod1 != -1);

    if (mod1 <= 0) {
        for (int i = 0; i < n; ++i) {
            if (s[i] == '1') {
                cout << 0 << endl;
            }
            else cout << 1 << endl;
        }
        return;
    }

    vec<int> s1(n + 1), s2(n + 1), s3(n + 1);
    vec<int> p1(n + 1), p2(n + 1), p3(n + 1);
    if (s[0] == '1') {
        p1[n - 1] = bin_pow(2, n - 1, mod1);
        p2[n - 1] = bin_pow(2, n - 1, mod2);
        p3[n - 1] = bin_pow(2, n - 1, mod3);
    }
    for (int i = 1; i < n; ++i) {
        p1[n - i - 1] = p1[n - i];
        if (s[i] == '1') {
            p1[n - i - 1] = (p1[n - i - 1] + bin_pow(2, n - i - 1, mod1)) % mod1;
        }

        p2[n - i - 1] = p2[n - i];
        if (s[i] == '1') {
            p2[n - i - 1] = (p2[n - i - 1] + bin_pow(2, n - i - 1, mod2)) % mod2;
        }

        p3[n - i - 1] = p3[n - i];
        if (s[i] == '1') {
            p3[n - i - 1] = (p3[n - i - 1] + bin_pow(2, n - i - 1, mod3)) % mod3;
        }
    }

    if (s[n - 1] == '1') {
        s1[0] = 1 % mod1;
        s2[0] = 1 % mod2;
        s3[0] = 1 % mod3;
    }
    for (int i = n - 2; i >= 0; --i) {
        s1[n - i - 1] = s1[n - i - 2];
        if (s[i] == '1') {
            s1[n - i - 1] = (s1[n - i - 1] + bin_pow(2, n - i - 1, mod1)) % mod1;
        }

        s2[n - i - 1] = s2[n - i - 2];
        if (s[i] == '1') {
            s2[n - i - 1] = (s2[n - i - 1] + bin_pow(2, n - i - 1, mod2)) % mod2;
        }

        s3[n - i - 1] = s3[n - i - 2];
        if (s[i] == '1') {
            s3[n - i - 1] = (s3[n - i - 1] + bin_pow(2, n - i - 1, mod3)) % mod3;
        }
    }

    for (int i = 0; i < n; ++i) {
        int p = n - i - 1;
        int sum = 0;
        if (s[i] == '0') {
            sum = (bin_pow(2, p, mod3) + (p + 1 == n ? 0 : p3[p + 1]) + (p - 1 >= 0 ? s3[p - 1] : 0)) % mod3;
        }
        else {  
            sum = ((p + 1 == n ? 0 : p1[p + 1]) + (p - 1 >= 0 ? s1[p - 1] : 0)) % mod1;
        }
        int ans = 1; 
        int x = sum;
        while (x > 0) {
            ans++;
            x %= __builtin_popcount(x);
        }
        cout << ans << endl;
    }
}