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()

X_pc = X.count("1")
X_res_m1 = 0
X_res_p1 = 0
for i, s in enumerate(reversed(X)):
    if s == "1":
        if X_pc > 1:
            X_res_m1 += pow(2, i, X_pc - 1)
            X_res_m1 %= X_pc - 1

        X_res_p1 += pow(2, i, X_pc + 1)
        X_res_p1 %= X_pc + 1

f = [-1] * (X_pc + 1)
f[0] = 0
for n in range(1, X_pc + 1):
    pc = format(n, "b").count("1")
    f[n] = f[n % pc] + 1

ans = []
for i, s in enumerate(reversed(X)):
    if s == "1":
        if X_pc > 1:
            fx = (X_res_m1 - pow(2, i, X_pc - 1)) % (X_pc - 1)
            ans.append(f[fx] + 1)
        else:
            ans.append(0)

    else:
        fx = (X_res_p1 + pow(2, i, X_pc + 1)) % (X_pc + 1)
        ans.append(f[fx] + 1)

print(*reversed(ans), sep="\n")