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: Runtime Error)

n = int(input())
x = list(input())
y = int(''.join(x), 2)

memo = [0]


def popcount(x):
    '''xの立っているビット数をカウントする関数
    (xは64bit整数)'''

    # 2bitごとの組に分け、立っているビット数を2bitで表現する
    x = x - ((x >> 1) & 0x5555555555555555)

    # 4bit整数に 上位2bit + 下位2bit を計算した値を入れる
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)

    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f  # 8bitごと
    x = x + (x >> 8)  # 16bitごと
    x = x + (x >> 16)  # 32bitごと
    x = x + (x >> 32)  # 64bitごと = 全部の合計
    return x & 0x0000007f


for i in range(1, n):
    tmp = i % popcount(i)
    memo.append(memo[tmp] + 1)

B = 2**(n-1)
for d in x:
    tmp = y
    if d == '0':
        tmp += B
    else:
        tmp -= B

    tmp = int(tmp)

    if tmp == 0:
        print(0)
        B = B // 2
        continue

    tmp = tmp % popcount(tmp)
    print(memo[tmp] + 1)
    B = B // 2