#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
long long a[200002];
int b[200002] = {0};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] == 0) v.push_back(i);
  }
  memset(b, 1e6, sizeof b);
  for (int k = 0; k < v.size(); ++k) {
    b[v[k]] = 0;
    for (int i = v[k] - 1, j = 1; i >= 0 && a[i] != 0; --i, ++j)
      b[i] = min(b[i], j);
    for (int i = v[k] + 1, j = 1; i < n && a[i] != 0; i++, j++)
      b[i] = min(b[i], j);
  }
  for (int i = 0; i < n; ++i) cout << b[i] << ' ';
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
long long a[200002];
int b[200002] = {0};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] == 0) v.push_back(i);
  }
  memset(b, 1e6, sizeof b);
  for (int k = 0; k < v.size(); ++k) {
    b[v[k]] = 0;
    for (int i = v[k] - 1, j = 1; i >= 0 && a[i] != 0; --i, ++j)
      b[i] = min(b[i], j);
    for (int i = v[k] + 1, j = 1; i < n && a[i] != 0; i++, j++)
      b[i] = min(b[i], j);
  }
  for (int i = 0; i < n; ++i) cout << b[i] << ' ';
}

```