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
readline = sys.stdin.readline

N = int(readline())
X = readline().rstrip()

X0 = int(X, 2)
count1 = X.count("1")

X0mod0to1 = X0 % (count1 + 1) # 0->1に変化した際の、X0を割った余り
if count1 > 1:
  X0mod1to0 = X0 % (count1 - 1) # 1->0に変化した際の、X0を割った余り
else:
  X0mod1to0 = 0

from collections import defaultdict
dp = defaultdict(int)
dp[0] = 0
def f(x): # 愚直にf(x)を求める式
  if x in dp:
    return dp[x]
  pc = bin(x).count("1")
  val = f(x % pc) + 1
  dp[x] = val
  return val
  
val0to1 = 1
val1to0 = 1
ans = [-1] * N

for i in range(N - 1, -1, -1): # 下の桁から順に、f(Xi)の値を求める
  if X[i] == "0": # 0を1に変える場合
    pat = (X0mod0to1 + val0to1 % (count1 + 1)) % (count1 + 1)
    ans[i] = f(pat) + 1
  elif X[i] == "1": # 1を0に変える場合
    if count1 > 1:
      pat = (X0mod1to0 - val1to0 % (count1 - 1)) % (count1 - 1)
      ans[i] = f(pat) + 1
    else:
      ans[i] = 0
  val0to1 = val0to1 * 2 % (count1 + 1)
  if count1 != 1:
    val1to0 = val1to0 * 2 % (count1 - 1)

print(*ans, sep = "\n")