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: Runtime Error)

import sys

sys.setrecursionlimit(10 ** 7)

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

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

pc = X.count("1")

# X == 0 の時
if pc == 0:
    for i in range(N):
        print(1)
    exit()

sml = pc - 1
big = pc + 1

b = 0
mod_s = 0
mod_b = 0
for x in X[::-1]:
    if x == "1":
        mod_s += pow(2, b, sml)
        mod_s %= sml

        mod_b += pow(2, b, big)
        mod_b %= big

    b += 1

# print(mod_s, mod_b, sml, big)


ELE = []

b = 0
for x in X[::-1]:
    if x == "0":
        t = (mod_b + pow(2, b, big)) % big
        ELE.append((t, 1))
    else:
        t = (mod_s - pow(2, b, sml)) % sml
        ELE.append((t, 1))
    b += 1


def solve(x, c):
    if x == 0:
        return c

    b = 0
    nx = x
    while nx > 0:
        b += nx & 1
        nx >>= 1

    ret = solve(x % b, c + 1)
    return ret


ANS = []
for x, pc in ELE:
    ANS.append(solve(x, pc))

print("\n".join(map(str, ANS[::-1])))