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: Accepted)

N = int(input())
X = input()
def popcount(n):
    tmp = n
    cnt = 0
    while tmp:
        cnt += tmp % 2
        tmp >>= 1
    return cnt
def f(n):
    tmp = n
    cnt = 0
    while tmp:
        cnt += 1
        ff = popcount(tmp)
        tmp %= ff
    return cnt
cnt = 0
LL = [0 for _ in range(200010)]
for i in range(200010):
    LL[i] = f(i)
for i in range(N):
    cnt += X[i] == '1'
A = [1 for _ in range(N+1)]
B = [1 for _ in range(N+1)]
for i in range(N):
    A[i+1] = A[i] << 1
    A[i+1] %= cnt + 1
for i in range(N):
    B[i+1] = B[i] << 1
    if cnt > 1:
        B[i+1] %= cnt - 1
    else:
        B[i+1] = 0
S = 0
T = 0
for i in range(N):
    S += X[i] == '1'
    T += X[i] == '1'
    if i != N-1:
        S <<= 1
        T <<= 1
        S %= cnt + 1
        if cnt > 1:
            T %= cnt - 1
        else:
            T = 0
C = [S for _ in range(N)]
D = [T for _ in range(N)]
for i in range(N):
    if X[i] == '1':
        D[i] -= B[N-1-i]
        if cnt == 1:
            print(0)
        else:
            D[i] %= cnt - 1
            print(LL[D[i]]+1)
    else:
        C[i] += A[N-1-i]
        C[i] %= cnt + 1
        print(LL[C[i]]+1)