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 = input()
original_pop_count = x.count('1')
one_pop_count = original_pop_count-1
zero_pop_count = original_pop_count+1

one_mod = 0
zero_mod= 0
for b in x:
    if one_pop_count!=0:
        one_mod=(one_mod*2+int(b))%one_pop_count
    zero_mod=(zero_mod*2+int(b))%zero_pop_count
#予めf[i]を計算する
f = [0]*220000
pop_count = [0]*220000
for i in range(1,220000):
    pop_count[i]= pop_count[i//2] + i%2
    f[i] = f[i%pop_count[i]]+1
one_pow2 = [1]*220000
zero_pow2 = [1]*220000
for i in range(1,220000):
    if one_pop_count!=0:
        one_pow2[i] = one_pow2[i-1]*2%one_pop_count
    zero_pow2[i] = zero_pow2[i-1]*2%zero_pop_count


for i in range(n-1,-1,-1):
    if x[n-1-i]=='1':
        if one_pop_count!=0:
            nxt = one_mod
            nxt-=one_pow2[i]
            nxt%=one_pop_count
            print(f[nxt]+1)
        else:
            print(0)
    if x[n-1-i]=='0':
        nxt = zero_mod
        nxt+=zero_pow2[i]
        nxt%=zero_pop_count
        print(f[nxt]+1)