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)



def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    mod=10**9+7
    def popcount(x):
        '''xの立っているビット数をカウントする関数
        (xは64bit整数)'''

        # 2bitごとの組に分け、立っているビット数を2bitで表現する
        x = x - ((x >> 1) & 0x5555555555555555)

        # 4bit整数に 上位2bit + 下位2bit を計算した値を入れる
        x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)

        x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f # 8bitごと
        x = x + (x >> 8) # 16bitごと
        x = x + (x >> 16) # 32bitごと
        x = x + (x >> 32) # 64bitごと = 全部の合計
        return x & 0x0000007f
    
    N=I()
    Xst=input()
    
    dp=[0]*(N+1)
    for i in range(1,N+1):
        X=i
        X=X%popcount(X)
        dp[i]=dp[X]+1 
        
    a=Xst.count("1")
    X=int(Xst,2)
    
    
    
    for i in range(N):
        if Xst[i]=="0":
            xa=a+1
            X2=X+(1<<(N-i-1))
        else:
            xa=a-1
            X2=X-(1<<(N-i-1))
        cnt=0
        #print(X2,xa)
        if X2!=0:
            X2=X2%xa
            cnt=dp[X2]+1
        print(cnt)
    
    
        

main()