You are given the array of integer numbers a0, a1, ..., an - 1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — length of the array a. The second line contains integer elements of the array separated by single spaces ( - 109 ≤ ai ≤ 109).

Output

Print the sequence d0, d1, ..., dn - 1, where di is the difference of indices between i and nearest j such that aj = 0. It is possible that i = j.

Examples

Input

9
2 1 0 3 0 0 3 2 4


Output

2 1 0 1 0 0 1 2 3 

Input

5
0 1 2 3 4


Output

0 1 2 3 4 

Input

7
5 6 0 1 -2 3 4


Output

2 1 0 1 2 3 4 
/*
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 * and open the template in the editor.
 */

/**
 *
 * @author As.Yasin
 */
import java.util.*;

public class DistancesToZeroB {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        StringBuilder out = new StringBuilder();
        int n = in.nextInt();
        int noOfZeros = 0;
        int[] arr = new int[n];
        int[] zeros = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
            if(arr[i] == 0){
                zeros[noOfZeros++] = i;
            }
        }
        int j = 0;
        for (int i = 0; i < arr.length; i++) {
            if(zeros[j] >= i){
                out.append(Math.abs(i-zeros[j])+" ");
            }
            else if(j+1 == noOfZeros){
                out.append(Math.abs(i-zeros[j])+" ");
            }
            else{
                if(Math.abs(i-zeros[j]) < Math.abs(i-zeros[j+1])){
                    out.append(Math.abs(i-zeros[j])+" ");
                }
                else{
                    out.append(Math.abs(i-zeros[++j])+" ");
                }
            }
        }
        System.out.println(out.deleteCharAt(out.length()-1));
    }
}
