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: Accepted)

#include <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#include <bitset>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a); i > (b); i--)
#define ALL(a) (a).begin(), (a).end()

int popcount(int n) {
    int ret = 0;
    while (n > 0) {
        if ((n & 1) == 1) ret++;
        n >>= 1;
    }
    return ret;
}

int f(int x) {
    if (x == 0) return 0;
    return f(x % popcount(x)) + 1;
}

int main() {
    int N;
    cin >> N;
    string X;
    cin >> X;

    bitset<200000> target(X);
    vector<int> ans(N);
    int initial_pop = target.count();
    REP(b, 2) {
        int new_pop = b == 0 ? initial_pop + 1 : initial_pop - 1;
        if (new_pop <= 0) continue;

        int base = 0;
        RREP(i, N) {
            base = (base << 1) + target.test(i);
            base %= new_pop;
        }
        int diff = 1;
        REP(i, N) {
            if (target.test(i) == b) {
                int r = b == 0 ? (base + diff) : (base - diff + new_pop);
                r %= new_pop;
                ans[N - i - 1] = f(r) + 1;
            }
            diff = (diff << 1) % new_pop;
        }
    }
    REP(i, N) cout << ans[i] << endl;
    return 0;
}