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 = str(input())


def power(a, n, mod):
    bi=str(format(n,"b")) #2進数
    res=1
    for i in range(len(bi)):
        res=(res*res) %mod
        if bi[i]=="1":
            res=(res*a) %mod
    return res

p = x.count('1')

if p == 0:
    ans = [1]*n
    for i in range(n):
        print(ans[i])
    exit()

if p == 1:
    if x[-1] != '1':
        ans = [1]*n
        for i in range(n):
            if x[i] == '1':
                ans[i] = 0
        ans[-1] = 2
    else:
        ans = [2]*n
        ans[-1] = 0
    for i in range(n):
        print(ans[i])
    exit()

x = list(x)
x.reverse()
temp1 = 0
for i in range(len(x)):
    temp1 += int(x[i])*power(2, i, p+1)
temp1 %= (p+1)

temp2 = 0
for i in range(len(x)):
    temp2 += int(x[i])*power(2, i, p-1)
temp2 %= (p-1)

ans = [0]*n

for i in range(n):
    if x[n-1-i] == '1':
        temp = temp2-power(2, n-1-i, p-1)
        c = p-1
    else:
        temp = temp1+power(2, n-1-i, p+1)
        c = p+1
    temp %= c
    #print(i, temp, c)
    cnt = 1
    while temp != 0:
        x_ = format(temp, 'b')
        c = x_.count('1')
        x_ = list(x_)
        x_.reverse()
        temp = 0
        for j in range(len(x_)):
            temp += int(x_[j])*power(2, j, c)
        temp %= c
        cnt += 1
    #print(i, cnt)
    ans[i] = cnt

for i in range(n):
    print(ans[i])