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: Runtime Error)

n=int(input())

s=input()
X=int(s,2)

one_cnt=s.count("1")

modp=[0]*n
modm=[0]*n

for i in range(n):
    modp[i]=(pow(2,(n-i-1),one_cnt+1)*int(s[i]))%(one_cnt+1)
    modm[i]=(pow(2,(n-i-1),one_cnt-1)*int(s[i]))%(one_cnt-1)

ans=[]
memo_one=[0]*(n+1)

#print(modp,modm)

for i in range(n):
    cnt=0
    mod=0

    for j in range(n):
        if s[i]=="1":
            if i!=j:
                mod=(mod+modm[j])%(one_cnt-1)
        else:
            if i!=j:
                mod=(mod+modp[j])%(one_cnt+1)
            else:
                mod=(mod+2**(n-i-1))%(one_cnt+1)
    cnt=1

   # print(i,mod)

    if mod==0:
        ans.append(cnt)
    else:
        while 1:
            cnt+=1
            if memo_one[mod]!=0:
                mod=mod%(memo_one[mod])
            else:
                tmp=(bin(mod).count("1"))
                memo_one[mod]=tmp
                mod=mod%(tmp)
            if mod==0:
                ans.append(cnt)
                break
            #print(mod,one_cnt,s[i],cnt)




#print(ans_memo)

for item in ans:
    print(item)