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: Wrong Answer)

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

def calc(S, L):
    cnt = 0
    for i in range(L):
        if S[i] == "1":
            cnt += 1

    a_1 = 1
    r_1 = [0] * (L+1)
    for i in reversed(range(L)):
        r_1[i] = (r_1[i+1] + int(S[i]) * a_1) % cnt
        a_1 = (a_1 * 2) % cnt

    return r_1[0]

def calc_first(S, L):
    cnt = 0
    for i in range(L):
        if S[i] == "1":
            cnt += 1

    a_1 = 1
    a_2 = 1
    r_1 = [0] * (L+1)
    r_2 = [0] * (L+1)
    for i in reversed(range(L)):
        r_1[i] = (r_1[i+1] + int(S[i]) * a_1) % (cnt+1)
        if cnt-1 > 0:
            r_2[i] = (r_2[i+1] + int(S[i]) * a_2) % (cnt-1)

        a_1 = (a_1 * 2) % (cnt + 1)
        if cnt-1 > 0:
            a_2 = (a_2 * 2) % (cnt - 1)

    for i in range(L):
        if S[i] == "0":
            ans = (r_1[0] - r_1[i] + r_1[i+1] + pow(2, N-i-1, (cnt+1))) % (cnt+1)
        else:
            if cnt - 1 == 0:
                ans = 0
            else:
                ans = (r_2[0] - r_2[i] + r_2[i+1]) % (cnt-1)

        cnt_ans = 1
        while ans > 0:
            ans = calc(str(bin(ans))[2:], len(str(bin(ans))[2:]))
            cnt_ans += 1
        print(cnt_ans)

calc_first(X, N)