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)

import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
read = sys.stdin.buffer.read
sys.setrecursionlimit(10 ** 7)
INF = float('inf')

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

# N = 2 * (10 ** 5)
# X = format(1 << (N - 1), "b")

from functools import lru_cache

@lru_cache(maxsize=None)
def F2(Xi):
    popcount = Xi.count("1")
    if popcount == 0:
        return True, Xi
    Xi = eval("0b" + Xi) % popcount
    Xi = format(Xi, "b")
    return False, Xi

@lru_cache(maxsize=None)
def F(Xi):
    ans = 0
    # print("start Xi: {} {}".format(eval("0b" + Xi), Xi))
    while True:
        is_end, Xi = F2(Xi)
        if is_end:
            break
        ans += 1
    return ans

# for i in range(1000, 1001):
#     ans = F(format(i, "b"))
#     print(ans)
# quit()

# print(bin(X))
for i in range(1, N+1):
    keta = N - i
    # print("keta: {}".format(keta))
    # Xi = list(X)
    # if X[keta] == "1":
    #     Xi[keta] = "0"
    # else:
    #     Xi[keta] = "1"
    # Xi = "".join(Xi)
    Xi = format(eval("0b" + X) ^ 1 << keta, "b")
    # print("Xi: {} {}".format(eval("0b" + Xi), Xi))
    ans = F(Xi)
    print(ans)