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)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>

#define PI 3.14159265359
using namespace std;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define bit(n,k) (((ll)n>>(ll)k)&1) /*nのk bit目*/
#define pb push_back
#define eb emplace_back
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const long long INF= 1e+18+1;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll> >vvl;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> T;
const ll MOD=1000000007LL;
//const ll MOD=998244353LL;
//const ll MOD=1777777777LL;
//const ll MAX_V=114514LL;
//const ll MAX = 500010LL;
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
vl dx={0,0,1,-1};
vl dy={1,-1,0,0};

//素因数分解O(√n)
map<ll,ll>prime_factor(ll n){
  map<ll,ll>res;
  for(ll i=2;i*i<=n;i++){
    while(n%i==0){
      res[i]++;
      n/=i;
    }
  }
  if(n!=1)res[n]=1;
  return res;
}

const ll MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
//finvが階乗の逆元

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modpow(ll a, ll n, ll mod) {
  
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}



int main(){
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(10);
  /*--------------------------------*/
  
  ll n;cin>>n;
  string s;cin>>s;
  ll one=0;
  rep(i,n){
    if(s[i]=='1')one++;
  }
  vl pows_minus(n);
  vl pows_plus(n);
  pows_minus[0]=1,pows_plus[0]=1;
  for(ll i=1;i<n;i++){
    if(one>1)pows_minus[i]=pows_minus[i-1]*2%(one-1);
    pows_plus[i]=pows_plus[i-1]*2%(one+1);
  }
  vl x(n);
  ll x_plus=0,x_minus=0;
  for(ll i=0;i<n;i++){
    x_plus+=((s[n-1-i]-'0')*pows_plus[i])%(one+1);
    x_plus%=one+1;
    if(one>1)x_minus+=((s[n-1-i]-'0')*pows_minus[i])%(one-1);
    x_minus%=(one-1);
  }
  rep(i,n){
    if(s[i]=='1'){
      if(one==1){
        cout<<0<<endl;
        continue;
      }
      ll nx=(x_minus+(one-1-pows_minus[n-1-i]))%(one-1);
      ll ans=1;
      while(nx>0){
        nx=(nx%__builtin_popcount(nx));
        ans++;
      }
      cout<<ans<<endl;
    }
    if(s[i]=='0'){
      //cout<<x_plus<<" "<<pows_minus[n-1-i]<<" "<<one+1<<endl;
      ll nx=(x_plus+pows_plus[n-1-i])%(one+1);
      ll ans=1;
      while(nx>0){
        nx=(nx%__builtin_popcount(nx));
        ans++;
      }
      cout<<ans<<endl;
    }
  }
  //cout<<x_plus<<endl;
}