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 
import java.util.Scanner;

public class Main {
	private static final int Max = (int) (2e5 + 10);
	private static int[] A;
	private static int d;
	private static int[] ans;
	private static int cnt;
	private static int n;

	public static void main(String[] args) {
		InitData();

//		for (int i = 0; i < cnt; i++) {
//			System.out.print(ans[i] + "  ");
//		}

		for (int i = 1; i <= n; i++) {
			d = EF(i);
			if (i == 1) {
				System.out.print(d);
			} else {
				System.out.print(" " + d);
			}
		}
		System.out.println();
	}

	private static void InitData() {
		Scanner cin = new Scanner(System.in);
		A = new int[Max];
		ans = new int[Max];
		cnt = 0;
		n = cin.nextInt();
		for (int i = 1; i <= n; i++) {
			A[i] = cin.nextInt();
			if (A[i] == 0) {
				ans[cnt++] = i;
			}
		}
	}

	// 二分查找，查找一个最接近K的值
	private static int EF(int k) {
		int start = 0;
		int end = cnt - 1;

		if (k > ans[end]) {
			return Math.abs(k - ans[end]);
		} else if (k < ans[start]) {
			return Math.abs(k - ans[start]);
		}
		while (start <= end) {
			int midle = (end - start) / 2 + start;
			if (ans[midle] == k) {
				return Math.abs(ans[midle] - k);
			} else if (ans[midle] > k) {
				end = midle - 1;
			} else {
				start = midle + 1;
			}
		}
		return Math.min(Math.abs(k - ans[start]), Math.abs(ans[end] - k));
	}

}
