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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define rrep(i,n) for(int i=(int)n-1;i>=0;--i)
using namespace std;
using ll = long long;
using P = pair<int, int>;
template<typename T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T> inline bool chmin(T& a,T b){if(b<a){a=b;return 1;}return 0;}
template<typename T> vector<T> make_vec(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>
auto make_vec(size_t a,Ts... ts){return vector<decltype(make_vec<T>(ts...))>(a,make_vec<T>(ts...));}
template<typename T,typename U,typename... V>
typename enable_if<is_same<T,U>::value>::type fill_v(U& u,const V... v){u=U(v...);}
template<typename T,typename U,typename... V>
typename enable_if<!is_same<T,U>::value>::type fill_v(U& u,const V... v){for(auto& e:u)fill_v<T>(e,v...);}

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    string s;
    cin >> n >> s;
    vector<int> x(n);
    rep(i, n) x[i] = s[i] - '0';
    int pc = 0;
    rep(i, n) pc += x[i];
    
    vector<int> ans(n);
    rep(b, 2) {
        int npc = pc;
        if (b == 0) npc++; else npc--;
        if (npc <= 0) continue;
        int r0 = 0;
        rep(i, n) {
            r0 = (r0 * 2) % npc;
            r0 += x[i];
        }
        int k = 1;
        rrep(i, n) {
            if (x[i] == b) {
                int r = r0;
                if (b == 0) r = (r + k) % npc; else r = (r - k + npc) % npc;
                ans[i] = f(r) + 1;
            }
            k = (k * 2) % npc;
        }
    }
    rep(i, n) cout << ans[i] << endl;
}