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)

#d 
n=int(input())
x=input()

def popcount(m):
    return bin(m).count("1")

intx = int( "0b"+x ,0 )
popcount0 = x.count("1")

intx_mod_pp1 = intx %(popcount0+1)
if popcount0!=1:
    intx_mod_pm1 = intx %(popcount0-1)
    
jisho = ["yet"]*n
for i in range(n):
    if x[i]=="1" and popcount0==1:
        print(0)
        continue
    
    pow2ip = pow(2,n-i-1,popcount0+1)
    if popcount0!=1:   pow2im = pow(2,n-i-1,popcount0-1)
    
    intxi=1
    c=0
    while intxi!=0:
        if c==0:
            if x[i]=="0":
                pop = popcount0+1
                intxi = (intx_mod_pp1 + pow2ip%pop )%pop
            else:
                pop = popcount0-1
                intxi = (intx_mod_pm1 - pow2im%pop )%pop        
        else:
            if jisho[intxi]=="yet":
                togo = intxi%popcount(intxi)
                jisho[intxi]= togo
                intxi = togo
            else:
                intxi =jisho[intxi]
        c+=1
    print(c)