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 
/**
 * Created by Martin on 4/28/2017.
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.PrintWriter;

public class Main {

    public static void main(String[]args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int N = Integer.parseInt(br.readLine());
        int[] a = new int[N];
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }
        int[] d = new int[N];
        int lz = Integer.MAX_VALUE;
        for (int i = 0; i < N; i++) {
            if (a[i]==0)lz = i;
            d[i] = (lz == Integer.MAX_VALUE) ? lz : Math.abs(i - lz);
        }
        lz = Integer.MAX_VALUE;
        for (int i = N-1; i >= 0; i--) {
            if (a[i]==0)lz = i;
            d[i] = Math.min(d[i], (lz == Integer.MAX_VALUE) ? lz : Math.abs(i - lz));
        }
        for (int i = 0; i < N; i++) {
            pw.print(d[i] + " ");
        }
        pw.println();
        pw.flush();
    }

    public static void debug(Object...args) {
        System.out.println(Arrays.deepToString(args));
    }
}
