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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50;
typedef long long ll;
ll qpow(ll a, ll b, int mod)
{
    if(mod == 0) return 0LL;
    ll ans = 1;a %= mod;
    for(ll i = b;i;i >>= 1,a = a * a % mod)
        if(i & 1) ans = ans * a % mod;
    return ans % mod;
}
int vis[maxn];//__builtin_popcount(h)
int gao(int x){
    int ans = 0;
    while(x){
        int now = __builtin_popcount(x);
        ans++;
        x %= now;
    }
    return ans;
}
void init()
{
    for(int i = 1;i <= 2e5;i++){
        vis[i] = gao(i);
        //if(i <= 20) cout << vis[i] << endl;
    }
}
int sum[maxn];
int main()
{
    std::ios::sync_with_stdio(false);
    int n;
    init();
    cin >> n;
    string s;
    cin >> s;
    int cnt1 = 0;
    reverse(s.begin(), s.end());
    for(int i = 0;i < s.size();i++){
        if(s[i] == '1') cnt1++;
    }
    int cnt2 = cnt1 + 1;
    cnt1 = cnt1 - 1;
    ll ans1 = 0, ans2 = 0;
    for(int i = 0;i < s.size();i++){
        if(s[i] == '1'){
            ans1 += qpow(2LL, i, cnt1);
            ans2 += qpow(2LL, i, cnt2);
        }
    }
    for(int i = 0;i < s.size();i++){
        if(s[i] == '1'){
            ll now = ans1 - qpow(2LL, i, cnt1);
            if(cnt1) now %= cnt1;
            sum[i] = vis[now] + 1;
        }
        else{
            ll now = ans2 + qpow(2LL, i, cnt2);
            if(cnt2)
            now %= cnt2;
            sum[i] = vis[now] + 1;
        }
    }
    for(int i = s.size() - 1;i >= 0;i--){
        cout << sum[i] << endl;
    }
    return 0;
}