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 Lua (LuaJIT 2.1.0) (judge verdict: Accepted)

local mfl, mce = math.floor, math.ceil
local n = io.read("*n", "*l")
local t = {}
local s = io.read()
local cnt1 = 0
for i = 1, n do
  t[i] = s:byte(i) == 49
  if t[i] then cnt1 = cnt1 + 1 end
end
local plusmap, minusmap = {}, {}
for i = 1, n do
  plusmap[i], minusmap[i] = 0, 0
end
do
  local dv = cnt1 + 1
  local mul = 1
  for i = n, 1, -1 do
    plusmap[i] = mul
    mul = (mul * 2) % dv
  end
end
if 2 < cnt1 then
  local dv = cnt1 - 1
  local mul = 1
  for i = n, 1, -1 do
    minusmap[i] = mul
    mul = (mul * 2) % dv
  end
end
local plus = 0
do
  local dv = cnt1 + 1
  for i = 1, n do
    if t[i] then
      plus = (plus + plusmap[i]) % dv
    end
  end
end
local minus = 0
if 1 < cnt1 then
  local dv = cnt1 - 1
  for i = 1, n do
    if t[i] then
      minus = (minus + minusmap[i]) % dv
    end
  end
end

local function firstdevide(p, dv)
  if t[p] then
    local currem = minus
    currem = (currem + dv - minusmap[p]) % dv
    return currem
  else
    local currem = plus
    currem = (currem + plusmap[p]) % dv
    return currem
  end
end

local function getpopcnt(x)
  local c = 0
  while 0 < x do
    if x % 2 == 1 then c = c + 1 end
    x = mfl(x / 2)
  end
  return c
end

for iq = 1, n do
  local dv = t[iq] and cnt1 - 1 or cnt1 + 1
  if dv == 0 then
    print(0)
  else
    local rem = firstdevide(iq, dv)
    local cnt = 1
    while 0 < rem do
      dv = getpopcnt(rem)
      rem = rem % dv
      cnt = cnt + 1
    end
    print(cnt)
  end
end