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 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
const int INF = 0x3f3f3f3f;
const double eps = 0.0000001;
const long long mod = 1e9 + 7;
long long mul(long long a, long long b) { return (a % mod * b % mod) % mod; }
long long pre(long long a, long long b) { return (a % mod / b % mod) % mod; }
long long n, m, k, x, y, z;
int a[N];
int vis[N];
vector<int> p;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0)
      vis[i] = -1, p.push_back(i);
    else
      vis[i] = INF;
  }
  for (int i = 0; i < p.size(); i++) {
    for (int j = p[i] - 1; j >= 1; j--) {
      if (vis[j] == -1)
        break;
      else {
        vis[j] = min(vis[j], p[i] - j);
        if (vis[j] != p[i] - j) break;
      }
    }
    for (int j = p[i] + 1; j <= n; j++) {
      if (vis[j] == -1)
        break;
      else {
        vis[j] = min(vis[j], j - p[i]);
        if (vis[j] != j - p[i]) break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == -1)
      cout << 0 << " ";
    else
      cout << vis[i] << " ";
  }
}
