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 = list(map(int, input()))
# N = 2*10**5
# X = list(map(int, "10"*(N//2)))

Xp = sum(X)
        
# print(Xp)
a = 0
b = 0
# c = 0
# :N
for i, x in enumerate(X):
    i_reverse = N-i-1
    if Xp > 1:
        a = (a + x * pow(2, i_reverse, Xp-1)) % (Xp-1)
    b = (b + x * pow(2, i_reverse, Xp+1)) % (Xp+1)
    # c = (c + x * (1<<i_reverse))
    # print(bin(c))

# print(a, b)

# :N
for i in range(N):
    i_reverse = N-i-1
    if Xp > 1:
        m = ([a, b][X[i]==0] + [-1, 1][X[i]==0] * pow(2, i_reverse, [Xp-1, Xp+1][X[i]==0])) % [Xp-1, Xp+1][X[i]==0]
    else:
        if X[i]==0:
            m = (b + pow(2, i_reverse, Xp+1)) % (Xp+1)
        else:
            print(0)
            continue
    count = 1
    # print(i, ([a, b][X[i]==0], [-1, 1][X[i]==0]*(1<<i_reverse)), [Xp-1, Xp+1][X[i]==0], X[i]==0)
    # print(m)
    while m > 0:
        p = sum([int(c) for c in str(bin(m)[2:])])
        m = m % p
        count += 1
    print(count)