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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys, math, itertools, collections, bisect
input = lambda: sys.stdin.buffer.readline().rstrip().decode('utf-8')
inf = float('inf') ;mod = 10**9+7
mans = inf ;ans = 0 ;count = 0 ;pro = 1

n = int(input())
X = str(input())
start_mod = X.count("1")
if start_mod == 1:
  for i in range(n):
    if X[i] == "1":
      key = i
else: key = -1

def bin_to_int(n,binstr,mod):
  retval = 0
  for i in range(n-1,-1,-1):
    retval += (pow(2,n-1-i,mod) if binstr[i] == "1" else 0)
    retval %= mod
  return retval

start_number_plus = bin_to_int(n,X,start_mod+1)
if key == -1:
  start_number_minus = bin_to_int(n,X,start_mod-1)
# print(start_number_plus,start_number_minus,"debug")

# print(key,"debug3")
for i in range(n):
  if i == key:print(0);continue
  if X[i] == "1":
    now_mod = start_mod - 1
    now_number = (start_number_minus - pow(2,n-1-i,now_mod))%now_mod
      
  else:
    now_mod = start_mod + 1
    now_number = (start_number_plus + pow(2,n-1-i,now_mod))%now_mod
  # print(i,now_number,now_mod,"debug")
  ci = 1
  while now_number != 0:
    # print(now_number,now_mod)
    now_mod = bin(now_number).count("1")
    now_number %= now_mod
    ci += 1
  print(ci)