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: Time Limit Exceeded)

from sys import stdin
import sys
import itertools
import array
import textwrap
import math

N = int(input())
X = input()
temp = X.count("1")
t = [0, 0, 0]

t[0] = int(X, 2) % (temp-1) if temp != 1 else 0
# t[1] = int(X, 2) % (temp)
t[2] = int(X, 2) % (temp+1)
# print(X)
cnt = 0
# X = int(X, 2)
# print(X)
for i in range(N):
    if X[i] == "1" and temp == 1:
        print(0)
        continue
    if X[i] == "1":
        p = (1 << (N-i-1)) % (temp-1)
        a = (t[0]-p) % (temp-1)
    elif X[i] == "0":
        m = (1 << (N-i-1)) % (temp+1)
        a = (t[2]+m) % (temp+1)
    cnt = 1
    # print(a)
    while a > 0:
        a = a % format(a, 'b').count("1")
        # print(a)
        cnt += 1
    print(cnt)