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: Accepted)

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

def e(x, y):
    t = [0] * len(x)
    if y <= 0: return (-1, t)
    if y == 1: return (0, t)
    res = 0
    z = 1
    for i in range(len(x) - 1, -1, -1):
        t[i] = z
        if x[i] == '1':
            res = (res + z) % y
        z = (z << 1) % y
    return (res, t)

def popcount(x):
    x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555)
    x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333)
    x = x + (x >> 4) & 0x0f0f0f0f0f0f0f0f
    x += x >> 8
    x += x >> 16
    x += x >> 32
    return x & 0x7f

fm = [-1] * (2 * (10 ** 5) + 1)
fm[0] = 0
def f(x):
    if fm[x] >= 0: return fm[x]
    fm[x] = 1 + f(x % popcount(x))
    return fm[x]


pc = X.count('1')
s0, t0 = e(X, pc + 1)
s1, t1 = e(X, pc - 1)

for i in range(N):
    t = 0
    if X[i] == '0':
        t = (s0 + t0[i]) % (pc + 1)
    else:
        if pc == 1:
            print(0)
            continue
        t = (s1 - t1[i]) % (pc - 1)
    print(1 + f(t))