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;
int a[200010];
unordered_set<int> s[200010];
int dp[200010];
bool ch[200010];
queue<int> och;
void f() {
  int n = och.front();
  och.pop();
  ch[n] = false;
  for (auto i : s[n]) {
    if (dp[i] == 0) {
      dp[i] = dp[n] + 1;
      ch[i] = true;
      och.push(i);
      continue;
    }
    if (dp[i] > dp[n] + 1) {
      dp[i] = dp[n] + 1;
      if (!ch[i]) {
        och.push(i);
        ch[i] = true;
      }
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  unordered_set<int> dp_1;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if ((i - a[i] >= 0) && (a[i - a[i]] % 2) == (a[i] % 2))
      s[i - a[i]].insert(i);
    if (i + a[i] < n && (a[i + a[i]] % 2) == (a[i] % 2)) s[i + a[i]].insert(i);
    if ((i - a[i] >= 0) && (a[i - a[i]] % 2) != (a[i] % 2)) dp[i] = 1;
    if (i + a[i] < n && (a[i + a[i]] % 2) != (a[i] % 2)) dp[i] = 1;
    if (dp[i]) dp_1.insert(i);
  }
  while (dp_1.size()) {
    och.push(*dp_1.begin());
    dp_1.erase(dp_1.begin());
    while (och.size()) f();
  }
  for (int i = 0; i < n; i++) {
    if (dp[i] == 0) dp[i] = -1;
    printf("%d ", dp[i]);
  }
  return 0;
}
