p02609 AIsing Programming Contest 2020 - Anything Goes to Zero
Let \mathrm{popcount}(n) be the number of `1`s 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).

Examples

Input

3
011


Output

2
1
1


Input

23
00110111001011011001110


Output

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3
```python
from sys import stdin
import sys
import itertools
import array
import textwrap
import math

N = int(input())
X = input()
temp = X.count("1")
t = [0, 0, 0]

t[0] = int(X, 2) % (temp-1) if temp != 1 else 0
t[2] = int(X, 2) % (temp+1)
cnt = 0
for i in range(N):
    if X[i] == "1" and temp == 1:
        print(0)
        continue
    if X[i] == "1":
        # p = (1 << (N-i-1)) % (temp-1)
        p = pow(2, N-1-i, temp-1)
        a = (t[0]-p) % (temp-1)
    elif X[i] == "0":
        # m = (1 << (N-i-1)) % (temp+1)
        m = pow(2, N-1-i, temp+1)
        a = (t[2]+m) % (temp+1)
    cnt = 1
    while a > 0:
        a = a % format(a, 'b').count("1")
        cnt += 1
    print(cnt)

```