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 PHP (7.4.4) (judge verdict: Accepted)

<?php
list($n) = ints();
list($x) = strs();
$pcx = popcount($x);
$mod2[0][$pcx + 1] = 1 % ($pcx + 1);
if ($pcx > 1) $mod2[0][$pcx - 1] = 1;
for ($i = 1; $i < $n; ++$i) {
    $mod2[$i][$pcx + 1] = $mod2[$i - 1][$pcx + 1] * 2 % ($pcx + 1);
    if ($pcx > 1) $mod2[$i][$pcx - 1] = $mod2[$i - 1][$pcx - 1] * 2 % ($pcx - 1);
}
$xmodp = f($x, $pcx + 1);
$xmodm = $pcx > 1 ? f($x, $pcx - 1) : 0;
for ($i = 0; $i < $n; ++$i) {
    if ($x[$i] === '0') {
        $xt = ($xmodp + $mod2[$n - 1 - $i][$pcx + 1]) % ($pcx + 1);
    } else {
        if ($pcx <= 1) {
            $ans[] = 0;
            continue;
        }
        $xt = ($xmodm - $mod2[$n - 1 - $i][$pcx - 1] + ($pcx - 1)) % ($pcx - 1);
    }
    $ans[] = f2($xt) + 1;
}
echo implode(PHP_EOL, $ans);
function f($x, $pc)
{
    global $n, $mod2;
    if ($pc === 0) return 0;
    $mod = 0;
    for ($i = 0; $i < $n; ++$i)
        if ($x[$i] === '1') $mod = ($mod2[$n - 1 - $i][$pc] + $mod) % $pc;
    return $mod;
}
function f2($n)
{
    if ($n === 0) return 0;
    return 1 + f2($n % popcount(decbin($n)));
}
function popcount($x)
{
    $res = 0;
    for ($i = 0; $i < strlen($x); ++$i)
        if ($x[$i] === '1') ++$res;
    return $res;
}
function strs()
{
    return explode(' ', trim(fgets(STDIN)));
}
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}