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: Runtime Error)

#!/usr/bin/env python3
N = int(input())
X = input()

x_num = int(X, 2)
cnt1 = X.count('1')

p_zero = x_num % (cnt1 + 1)
p_one = x_num % (cnt1 - 1)


def f(cnt):
    if (cnt > 0):
        pcount = p_count(cnt)
        return f(cnt % pcount) + 1
    else:
        return 0


def p_count(x):
    cnt = 0
    while (x > 0):
        if x % 2 != 0:
            cnt += 1
        x //= 2
    return cnt


def f1(x):
    cnt = 0
    while (x > 0):
        x %= bin(x).count('1')
        cnt += 1
    return cnt


for i, x in enumerate(list(X)):
    if (x == '1'):
        #x_num = int(''.join(X), 2)
        if (x_num - (2**(N-(i+1)))) != 0:
            print(f1((x_num - pow(2, (N - (i + 1)), cnt1 - 1)) % (cnt1 - 1)) + 1)
        else:
            print(0)
    else:
        #x_num = int(''.join(X), 2)
        print(f1((x_num + pow(2, (N - (i + 1)), cnt1 + 1)) % (cnt1 + 1)) + 1)