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)

// Hack it and have it ;) //
/*author* Priyanshu Shrivastav (from IIT Palakkad) *
 * *_ __ ___  _ ______ ___  _ ____   ___  ___| |_  *
 * | '_ ` _ \| '__/ __/ _ \| '_ \ \ / / |/ __| __| *
 * | | | | | | | | (_| (_) | | | \ V /| | (__| |_  *
 * |_| |_| |_|_|(_)___\___/|_| |_|\_/ |_|\___|\__| *
When I wrote this, only God and I understood what I was doing
 ** * * * * * * * Now, only God knows!* * * * * * */
#include      <bits/stdc++.h>
using namespace std;
#define IOS       ios_base::sync_with_stdio(false); cin.tie (nullptr)
#define PREC      cout.precision (10); cout << fixed
#ifdef CONVICTION
#include "/home/convict/Dropbox/myfiles/sport_coding/cplib/snippets/debug.h"
#else
#define debug(x...)
#endif
//Don’t practice until you get it right. Practice until you can’t get it wrong

const int Maxn = 1e6 + 10;
int dp[Maxn];

void preproc() {
}

int getVal (const string &s, int Mod) {
   int res = 0, x = 1;
   int sz = (int)s.size();
   for (int i = sz-1; i >= 0; --i) {
      if (s[i] == '1') {
         res = (res + x) % Mod;
      }
      x = x*2 % Mod;
   }
   return res;
}
int pw2Giveny (int y, int Mod) {
   int res = 1, x = 2;
   while (y) {
      if (y & 1) res = 1ll*res*x % Mod;
      x = 1ll*x*x % Mod;
      y >>= 1;
   }
   return res;
}

void solve() {
   int N;
   string X;
   cin >> N >> X;
   dp[0] = 0;
   for (int i = 1; i <= N; ++i) {
      dp[i] = 1 + dp[i % __builtin_popcount(i)];
   }

   int cntTotal = 0;
   for (char ch : X) cntTotal += (ch == '1');
   int modLow = cntTotal - 1, modHigh = cntTotal + 1;
   int lowVal = getVal(X, modLow);
   int highVal = getVal(X, modHigh);

   for (int i = 0; i < N; ++i) {
      int rem, firstStep;
      if (X[i] == '1') {
         if (cntTotal == 1) {
            cout << 0 << '\n';
            continue;
         }
         rem = pw2Giveny(N-i-1, modLow);
         firstStep = (lowVal - rem + modLow) % modLow;
      }
      else {
         rem = pw2Giveny(N-i-1, modHigh);
         firstStep = (highVal + rem) % modHigh;
      }
      cout << 1 + dp[firstStep] << '\n';
   }
}

signed main() {
  IOS; PREC;
  preproc();

  int tc = 1;
  // cin >> tc;
  for (int Tt = 1; Tt <= tc; ++Tt) {
    // cout << "Case #" << Tt << ": ";
    solve();
  }
  return EXIT_SUCCESS;
}