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 Python (3.8.2) (judge verdict: Runtime Error)

#!/usr/bin/env python

import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10**6)
INF = float("inf")

def popcount(n):
    ret = 0
    for i in range(18):
        if n>>i & 1:
            ret += 1
    return ret

def main():
    N = int(input())
    X = input().decode().rstrip()

    # X に含まれる "1" の数
    x_cnt = 0
    for i in range(N):
        if X[i] == "1":
            x_cnt += 1

    # i 桁目が 0 : x_cnt += 1
    # i 桁目が 1 : x_cnt -= 1
    # より, X mod x_cnt +1, X mod x_cnt-1 を事前計算しておく
    n_plus,n_minus = 0,0
    t_plus,t_minus = 1,1
    for i in range(N-1,-1,-1):
        if X[i]=="1":
            n_plus += t_plus
            n_plus %= x_cnt + 1
            n_minus += t_minus
            n_minus %= x_cnt - 1
        t_plus = (t_plus*2) % (x_cnt + 1)
        t_minus = (t_minus*2) % (x_cnt - 1)
    
    for i in range(N):
        ans = 1
        if X[i] == "1":
            # 1 -> 0 に変換した結果, １が全く存在しない場合
            #if x_cnt == 1:
            #    print(0)
            #    continue
            a = (n_minus - pow(2,(N-1-i), x_cnt-1)) % (x_cnt-1)
        else:
            a = (n_plus + pow(2,(N-1-i), x_cnt+1)) % (x_cnt+1)

        while a > 0:
            pop_cnt = popcount(a)
            a %= pop_cnt
            ans += 1

        print(ans)

if __name__ == "__main__":
    main()