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: Wrong Answer)

import sys
import itertools
# import numpy as np
import time
import math
import heapq
from collections import defaultdict
sys.setrecursionlimit(10 ** 7)
 
INF = 10 ** 18
MOD = 10 ** 9 + 7
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

# map(int, input().split())

N = int(input())
X = input()

def popcount(n):
    cnt = 0
    while n > 0:
        cnt += n & 1
        n //= 2
    return cnt
def f(n):
    cnt = 0
    while n != 0:
        n = n % popcount(n)
        cnt += 1
    return cnt

ones = X.count('1')
MOD1 = ones + 1
MOD2 = ones - 1
pow1 = [0] * (len(X) + 1)
pow2 = [0] * (len(X) + 1)
if MOD2 != 0:
    pow1[0] = 1
    pow2[0] = 1
    for i in range(1, len(X) + 1):
        pow1[i] = (pow1[i - 1] * 2) % MOD1
        pow2[i] = (pow2[i - 1] * 2) % MOD2

X1 = 0
X2 = 0
for i in range(N):
    if X[i] == '1':
        X1 += pow1[N - i - 1]
        X2 += pow2[N - i - 1]
X1 %= MOD1
if MOD2 != 0:
    X2 %= MOD2

for i in range(N):
    d = N - i - 1
    x = 0
    if X[i] == '0':
        x = (X1 + pow1[d]) % MOD1
        print(f(x) + 1)
    else:
        if MOD2 != 0:
            x = (X2 - pow2[d]) % MOD2
            print(f(x) + 1)
        else:
            print(0)