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 = 2*10**5
# X = "10"*10**5
n = int(input())
X = input()
c = 0
cnt = 0
for i,x in enumerate(X):
    c += (1<<(n-i-1))*int(x)
    cnt += int(x)

amari_cnt_p = c % (cnt+1)
if cnt != 1:
    amari_cnt_m = c % (cnt-1)
else:
    amari_cnt_m = 0

dp = []
for d in range(n):
    if X[d] == "0":
        amari = (amari_cnt_p + pow(2,n-d-1,cnt+1))%(cnt+1)
    else:
        if cnt != 1:
            amari = (2*(cnt-1) + amari_cnt_m - pow(2,n-d-1,cnt-1))%(cnt-1)
        else:
            amari = -1
    dp.append(amari)

memo = [-1]*(2*10**5+10)
memo[0] = 1
def calc(n):
    if n == -1:
        return 0
    if n == 0:
        return 1
    if memo[n] != -1:
        return memo[n]
    cnt = 2
    while True:
        c = bin(n).count("1")
        if n%c == 0:
            break
        n %= c
        cnt += 1
    memo[n] = cnt
    return cnt

for d in dp:
    ans = calc(d)
    print(ans)