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: Time Limit Exceeded)

import sys #追加

n=int(input())
sys.setrecursionlimit(10**6*2)

s=input()
X=int(s,2)
ans=[]

ans_memo=[0]*(10**6)
#print(ans_memo)
ans_memo[1]=1
ans_memo[2]=1

def DEF(x,cnt):
    global ans_memo
    #print(x,cnt,ans_memo)
    if x==0:
        return 0
    if x<10**6:    
        if ans_memo[x]!=0:
            return ans_memo[x]
        else:
            if x%(str(bin(x)).count("1"))==0:
                ans_memo[x]=1
                return 1
            else:
                tmp=DEF(x%(str(bin(x)).count("1")),cnt+1)+1
                ans_memo[x]=tmp
                return tmp
    else:
        if x%(str(bin(x)).count("1"))==0:
            return 1
        else:
            tmp=DEF(x%(str(bin(x)).count("1")),cnt+1)+1
            return tmp


ans=[]

for i in range(n):
    tmp=2**(n-i-1)
    tmp_num=X
    #1の時
    if tmp&tmp_num:
        tmp_num-=tmp
        #print(bin(tmp_num),n-i-1)
        ans.append(DEF(tmp_num,0))
        
    else:
        tmp_num+=tmp
        #print(bin(tmp_num),n-i-1)
        ans.append(DEF(tmp_num,0))

#print(ans_memo)

for item in ans:
    print(item)