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;
using i64 = int_fast64_t;
#define rep(i, N) for(i64 i = 0; i < N; i++)
#define Rep(i, k, N) for(i64 i = k; i < N; i++)
#define all(v) v.begin(), v.end()
#define eb emplace_back
#define Cout(s) cout << s << endl
string Yes(bool b) {if(b){return "Yes";} else {return "No";}}
string YES(bool b) {if(b){return "YES";} else {return "NO";}}

i64 f(i64& n){
    int cnt = 0;
    while(n){
        n %= __builtin_popcount(n);
        cnt++;
    }
    return cnt;
}

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

    vector<vector<i64>> POW2(2, vector<i64>(200001, 0)); // [2^k % (popcount(X) + 1),  2^k % (popcount(X) - 1)]
    vector<i64> XmPC(2, 0); // [X % (popcount(X) + 1),  X % popcount(X - 1)]

    i64 popcnt = count(X.begin(), X.end(), '1');
    POW2[0][0] = 1 % (popcnt + 1);
    if(popcnt - 1 > 0) POW2[1][0] = 1 % (popcnt - 1);

    for(int i = 0; i < N; i++){
        XmPC[0] = ((XmPC[0] << 1ll) + (X[i] - '0')) % (popcnt + 1);
        POW2[0][i+1] = (POW2[0][i] << 1ll) % (popcnt + 1);
        if(popcnt - 1 > 0) {
            XmPC[1] = ((XmPC[1] << 1ll) + (X[i] - '0')) % (popcnt - 1);
            POW2[1][i+1] = (POW2[1][i] << 1ll) % (popcnt - 1);
        }
    }

    for(int i = 0; i < N; i++){
        i64 b = (X[i] - '0');
        i64 p = (1 - 2*b);
        i64 n = (XmPC[b] + p*POW2[b][N-i-1]);
        while(n < 0) n += (popcnt + p);
        n %= (popcnt + p);
        cout << f(n) + 1 << endl;
    }

}