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 Java (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.math.*;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in) ;
		int n, bitCnt = 0, y, bit, bb, newBitCnt, rem, ans = 0, vx, vy ;
		String s ;
		char x ;
		BigInteger val = BigInteger.valueOf(0), newVal ;
		n = sc.nextInt() ;
		s = sc.next() ;
		
		for (int i = 0; i < n; i++) {
			x = s.charAt(i) ;
			y = x - '0' ;
			bb = n - 1 - i ;
			bit = (1 << bb) ;
			if (y == 1) {
				bitCnt++ ;
				val = val.add(BigInteger.valueOf(bit)) ;
			}
		}
		
		for (int i = 0; i < n; i++) {
			ans = 1 ;
			x = s.charAt(i) ;
			y = x - '0' ;
			bb = n - 1 - i ;
			bit = (1 << bb) ;
			if (y == 1) {
				newBitCnt = bitCnt - 1 ;
				newVal = val.subtract(BigInteger.valueOf(bit)) ;
			}
			else {
				newBitCnt = bitCnt + 1 ;
				newVal = val.add(BigInteger.valueOf(bit)) ;
			}
			newVal = newVal.remainder(BigInteger.valueOf(newBitCnt)) ;
			rem = newVal.intValue() ;
			while (rem != 0) {
				ans++ ;
				vx = rem ;
				vy = Integer.bitCount(vx) ;
				rem = vx % vy ;
			}
			System.out.println(ans) ;
		}
		System.gc(); 
	}
}