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>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)

ll modpow(ll a, ll n, ll modd = 1e9 + 7) {
    ll res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * a % modd;
        a = a * a % modd;
        n >>= 1;
    }
    return res;
}

int popcnt(int num) {
    int ans = 0;
    while (num > 0) {
        num &= num - 1;
        ans++;
    }
    return ans;
}

int filler(int num, int modd) {
    if (modd == 0) {
        return 0;
    }
    int res = num % modd;
    return filler(res, popcnt(res)) + 1;
}

int main() {
    int n;
    string x;
    cin >> n >> x;
    int basemod = 0;
    int onebase = 0;
    int zerobase = 0;
    rep(i, n) if (x[i] == '1') basemod++;
    if(basemod == 0){
        rep(i,n){
            cout << 0 << endl;
        }
        return 0;
    }
    rep(i, n) {
        if (x[i] == '1') {
            onebase += modpow(2, n - i - 1, basemod + 1);
            zerobase += modpow(2, n - i - 1, basemod - 1);
        }
    }
    rep(i, n) {
        int modthis = basemod;
        int toadd;
        if (x[i] == '1') {
            modthis--;
            if(modthis == 0){
                cout << 0 << endl;
                continue;
            }
            if(modthis == 1){
                cout << 1 << endl;
                continue;
            }
            toadd = modpow(2, n - i - 1, modthis);
            zerobase -= toadd;
            if (zerobase < 0) zerobase += modthis;
            cout << filler(zerobase, modthis) << endl;
            zerobase += toadd;
            if (toadd != 0)
                zerobase %= toadd;
        } else {
            modthis++;
            if(modthis == 1){
                cout << 1 << endl;
                continue;
            }
            toadd = modpow(2, n - i - 1, modthis);
            onebase += toadd;
            onebase %= modthis;
            cout << filler(onebase, modthis) << endl;
            onebase -= toadd;
            if(onebase < 0){
                onebase += modthis;
            }
        }
    }
}