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)

#include <bits/stdc++.h>
using namespace std;

#define CPP_STR(x) CPP_STR_I(x)
#define CPP_CAT(x,y) CPP_CAT_I(x,y)
#define CPP_STR_I(args...) #args
#define CPP_CAT_I(x,y) x ## y

using i8  = int8_t;
using u8  = uint8_t;
using i16 = int16_t;
using u16 = uint16_t;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;

using f32 = float;
using f64 = double;

#define FOR(i, start, end) for(i64 i = (start), CPP_CAT(i,xxxx_end)=(end); i < CPP_CAT(i,xxxx_end); ++i)
#define REP(i, n) FOR(i, 0, n)
#define REPS(i, n) for(i64 i = 1; i <= n; ++i)
#define RREP(i, n) for(i64 i = n - 1; i >= 0 ; --i)
#define RREPS(i, n) for(i64 i = n; i >= 1; --i)

#define ALL(f,c,...) (([&](decltype((c)) cccc) { return (f)(std::begin(cccc), std::end(cccc), ## __VA_ARGS__); })(c))

#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

#define pb push_back
#define mp make_pair

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; }

template<typename C>
i64 SIZE(const C& c) { return static_cast<i64>(c.size()); }

template<typename T, size_t N>
i64 SIZE(const T (&)[N]) { return static_cast<i64>(N); }


struct ProconInit {
    static constexpr int IOS_PREC = 15;
    static constexpr bool AUTOFLUSH = false;

    ProconInit() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(IOS_PREC);
        if(AUTOFLUSH)
            cout << unitbuf;
    }
} PROCON_INIT;


i64 popcount(i64 x) {
    i64 cnt = 0;
    while (x > 0) {
        cnt += x & 1;
        x >>= 1;
    }
    return cnt;
}


int main() {
    i64 N; cin >> N;
    string S; cin >> S;
    reverse(S.begin(), S.end());

    i64 X = 0;
    i64 p = 1;
    REP(i, N) {
        X += p * (S[i] - '0');
        p *= 2;
    }

    p /= 2;
    RREP(i, N) {
        i64 cnt = 0;
        i64 x = X ^ p;

        while (x > 0) {
            x = x % popcount(x);
            cnt++;
        }

        cout << cnt << endl;

        p /= 2;
    }

    return 0;
}