Breaking news from zombie neurology! It turns out that – contrary to previous beliefs – every zombie is born with a single brain, and only later it evolves into a complicated brain structure. In fact, whenever a zombie consumes a brain, a new brain appears in its nervous system and gets immediately connected to one of the already existing brains using a single brain connector. Researchers are now interested in monitoring the brain latency of a zombie. Your task is to write a program which, given a history of evolution of a zombie's nervous system, computes its brain latency at every stage.

Input

The first line of the input contains one number n – the number of brains in the final nervous system (2 ≤ n ≤ 200000). In the second line a history of zombie's nervous system evolution is given. For convenience, we number all the brains by 1, 2, ..., n in the same order as they appear in the nervous system (the zombie is born with a single brain, number 1, and subsequently brains 2, 3, ..., n are added). The second line contains n - 1 space-separated numbers p2, p3, ..., pn, meaning that after a new brain k is added to the system, it gets connected to a parent-brain <image>.

Output

Output n - 1 space-separated numbers – the brain latencies after the brain number k is added, for k = 2, 3, ..., n.

Example

Input

6
1
2
2
1
5


Output

1 2 2 3 4 
#include <bits/stdc++.h>
using namespace std;
long long n, par[200005][30], lv[200005], ans[200005], tmp;
long long lca(long long a, long long b) {
  if (lv[a] < lv[b]) swap(a, b);
  for (long long i = 19; i >= 0; i--)
    if (lv[a] - (1 << i) >= lv[b]) a = par[a][i];
  for (long long i = 19; i >= 0; i--)
    if (par[a][i] != par[b][i]) a = par[a][i], b = par[b][i];
  if (a != b) a = par[a][0];
  return a;
}
long long dist(long long a, long long b) {
  long long u = lca(a, b);
  return lv[a] + lv[b] - 2 * lv[u];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  lv[1] = 1;
  long long u = 1, v = 1;
  for (long long i = 2; i <= n; i++) {
    cin >> tmp;
    lv[i] = lv[tmp] + 1;
    par[i][0] = tmp;
    for (long long j = 1; j <= 19; j++) par[i][j] = par[par[i][j - 1]][j - 1];
    long long dist_1 = dist(i, u), dist_2 = dist(i, v), dist_3 = dist(u, v);
    if (dist_1 >= dist_2 && dist_1 >= dist_3)
      v = i;
    else if (dist_2 >= dist_3 && dist_2 >= dist_1)
      u = i;
    ans[i] = max(dist_1, max(dist_2, dist_3));
  }
  for (long long i = 2; i <= n; i++) cout << ans[i] << ' ';
}
