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>
#include<bit>
using namespace std;
#define rep(i,n) for (int i = 0; i < n; ++i)
using ll = long long;
const ll MOD = 1000000007;//998244353;
int main() {
   ios::sync_with_stdio(false);
   cin.tie(0);
   int n; cin >> n;
   vector<int> ans;
   string x; cin >> x;
   reverse(x.begin(),x.end());
   ll cnt = 0;
   rep(i,n)cnt += (x[i] == '1');
   vector<ll> mod2_up, mod2_down;
   mod2_up.push_back(1%(cnt+1));
   mod2_down.push_back(1%(cnt-1));
   rep(i,n-1){
      mod2_up.push_back((mod2_up[i]*2)%(cnt+1));
      mod2_down.push_back((mod2_down[i]*2)%(cnt-1));
   }
   ll modup = 0, moddown = 0;
   rep(i,n){
      if (x[i]=='1'){
         modup += mod2_up[i];
         moddown += mod2_down[i];
         modup %= (cnt+1);
         moddown %= (cnt-1);
      }
   }
   rep(i,n){
      int ansi = 1, newx;
      if (x[i] == '1'){
         newx = moddown;
         newx = (newx - mod2_down[i]+(cnt-1))%(cnt-1);
      }
      else{
         newx = modup;
         newx = (newx + mod2_up[i])%(cnt+1);
      }
      while (newx > 0)
      {
         int popcnt = 0;
         for (int i = newx; i > 0; i>>=1)
            popcnt += (i&1);
         newx %= popcnt;
         ansi++;
      }
      ans.push_back(ansi);
   }
   rep(i,n){
      cout << ans[n-i-1] << endl;
   }
}