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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
typedef long long ll;
ll modpow(ll a, ll n, ll MOD) {
	if (n == 0)
		return 1;
	if (n % 2 == 1)
		return (a * modpow(a, n - 1, MOD) % MOD);
	ll t = modpow(a, n / 2, MOD);
	return (t * t % MOD);
}
ll popCnt(ll num) {
	int res = 0;
	for (int i = 0; i < 64; i++) {
		if (num >> i & 1) {
			res++;
		}
	}
	return res;
}
int main() {
	int n;
	int cnt = 0;
	char x[200001];
	scanf("%d%s", &n, x);
	for (int i = 0; i < n; i++) {
		x[i] -= '0';
		if (x[i] == 1) {
			cnt++;
		}
	}
	ll mx = 0, px = 0;
	bool is0 = false;
	for (int i = 0; i < n; i++) {
		px += x[i] * modpow(2, n - i - 1, cnt + 1) % (cnt + 1);
		if (cnt == 1) {
			is0 = true;
		} else {
			mx += x[i] * modpow(2, n - i - 1, cnt - 1) % (cnt - 1);
		}
	}
	for (int i = 0; i < n; i++) {
		ll now = 0;
		if (x[i] == 1) {
			if (is0) {
				printf("0\n");
				continue;
			}
			now = mx;
			now += (cnt - 1);
			now -= modpow(2, n - i - 1, cnt - 1);
			now %= (cnt - 1);
		} else {
			now = px;
			now += modpow(2, n - i - 1, cnt + 1);
			now %= (cnt + 1);
		}
		int ans = 1;
		while (popCnt(now) != 0) {
			// printf("x:%d", num);
			int c = popCnt(now);
			// printf(" cnt:%d\n", c);
			now = now % c;
			ans++;
		}
		printf("%d\n", ans);
	}
	return 0;
}