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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(){return +cin[cid++];}
function nextstr(){return cin[cid++];}
function nextbig(){return BigInt(cin[cid++]);}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}

var myOut = main();
if(myOut !== undefined)console.log(String(myOut));

function main(){
  var n = next();
  var x = nextstr();
  var bc = 1;
  for(var i = 0; i < n; i++)if(x[i] === "1")bc++;
  var ans = xArray(0,n);
  var m1 = 0;
  for(var i = 0; i < n; i++){
    m1 *= 2;
    m1 += +x[i];
    m1 %= bc;
  }
  var c = 1;
  for(var i = n; i--;){
    if(x[i] === "0"){
      ans[i] = 1;
      var t = m1+c;
      t %= bc;
      while(t){
        t %= bitcount(t);
        ans[i]++;
      }
    }
    c = c*2%bc;
  }
  bc -= 2;
  if(bc < 1)return ans.join("\n");
  m1 = 0;
  for(var i = 0; i < n; i++){
    m1 *= 2;
    m1 += +x[i];
    m1 %= bc;
  }
  c = bc-1;
  for(var i = n; i--;){
    if(x[i] === "1"){
      ans[i] = 1;
      var t = m1+c;
      t %= bc;
      while(t){
        t %= bitcount(t);
        ans[i]++;
      }
    }
    c = c*2%bc;
  }
  return ans.join("\n");
}

function bitcount(bits){
  bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
  bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
  bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
  bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
  return (bits & 0x0000ffff) + (bits >>16 & 0x0000ffff);
}