You are given an array a consisting of n integers. In one move, you can jump from the position i to the position i - a_i (if 1 ≤ i - a_i) or to the position i + a_i (if i + a_i ≤ n).

For each position i from 1 to n you want to know the minimum the number of moves required to reach any position j such that a_j has the opposite parity from a_i (i.e. if a_i is odd then a_j has to be even and vice versa).

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.

Output

Print n integers d_1, d_2, ..., d_n, where d_i is the minimum the number of moves required to reach any position j such that a_j has the opposite parity from a_i (i.e. if a_i is odd then a_j has to be even and vice versa) or -1 if it is impossible to reach such a position.

Example

Input


10
4 5 7 6 7 5 4 4 6 4


Output


1 1 1 2 -1 1 1 3 1 1 
#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
using ll = long long;
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = (int)0; i < (int)n; ++i) cin >> a[i];
  vector<int> answer(n, -1);
  vector<vector<int>> rev_edges(n);
  vector<int> q;
  for (int i = (int)0; i < (int)n; ++i) {
    for (int j : {i - a[i], i + a[i]}) {
      if (j >= 0 && j < n) {
        if (a[i] % 2 != a[j] % 2) {
          answer[i] = 1;
          q.push_back(i);
        }
        rev_edges[j].push_back(i);
      }
    }
  }
  for (int z = (int)0; z < (int)q.size(); ++z) {
    int i = q[z];
    for (int j : rev_edges[i]) {
      if (answer[j] == -1) {
        answer[j] = answer[i] + 1;
        q.push_back(j);
      }
    }
  }
  for (int x : answer) cout << x << " ";
  cout << endl;
  return 0;
}
