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>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int countSetBit(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
long long mod = 1ll * 1e9 + 7;
const int N = 1000100;
const int M = 6;
const double eps = 1e-6;
using namespace std;
long long a[N];
int l[N], r[N];
int main() {
  int n = nextInt();
  for (int i = 0; i < n; i++) {
    a[i] = nextLong();
  }
  int left = -1;
  for (int i = 0; i < n; i++) {
    if (a[i] == 0) {
      left = i;
    }
    l[i] = left;
  }
  int right = -1;
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == 0) {
      right = i;
    }
    r[i] = right;
  }
  for (int i = 0; i < n; i++) {
    int x = l[i];
    int y = r[i];
    int ans = 10000000;
    if (x != -1) {
      ans = i - x;
    }
    if (y != -1) {
      ans = min(ans, y - i);
    }
    cout << ans << " ";
  }
  return 0;
}
