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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  std::cin >> n;
  std::vector<std::vector<int>> edges(n);
  std::vector<int> a(n);
  for (int i = 0; i < n; i++) {
    std::cin >> a[i];
    if (i - a[i] >= 0) {
      edges[i - a[i]].push_back(i);
    }
    if (i + a[i] < n) {
      edges[i + a[i]].push_back(i);
    }
  }
  const int INF = 1e9;
  std::vector<int> ans(n, INF);
  for (int rep = 0; rep < 2; rep++) {
    std::vector<int> dist(n, INF);
    std::queue<int> q;
    for (int i = 0; i < n; i++) {
      if (a[i] % 2 == rep) {
        dist[i] = 0;
        q.push(i);
      }
    }
    while (!q.empty()) {
      int on = q.front();
      q.pop();
      for (auto to : edges[on]) {
        if (dist[to] > 1 + dist[on]) {
          dist[to] = 1 + dist[on];
          q.push(to);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (a[i] % 2 != rep) {
        ans[i] = dist[i] == INF ? -1 : dist[i];
      }
    }
  }
  for (int i = 0; i < n; i++) {
    std::cout << ans[i] << (i + 1 == n ? '\n' : ' ');
  }
}
