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;
const int N = 200001;
int n, par[N][20], lvl[N];
int lca(int u, int v) {
  if (lvl[u] < lvl[v]) swap(u, v);
  for (int i = 19; i >= 0; --i)
    if (par[u][i] != -1 && lvl[par[u][i]] >= lvl[v]) u = par[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; --i)
    if (par[u][i] != -1 && par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  return par[u][0];
}
int dis(int u, int v) { return lvl[u] + lvl[v] - 2 * lvl[lca(u, v)]; }
int main() {
  cin >> n;
  memset(par, -1, sizeof(par));
  int u = 1, v = 0, res = 0;
  lvl[1] = 1;
  for (int i = 2; i <= n; ++i) {
    cin >> par[i][0];
    lvl[i] = lvl[par[i][0]] + 1;
    for (int j = (1), _b = (19); j <= _b; ++j)
      if (par[i][j - 1] != -1)
        par[i][j] = par[par[i][j - 1]][j - 1];
      else
        break;
    if (v == 0) {
      v = i;
      res = dis(u, v);
    } else {
      int d1 = dis(v, i), d2 = dis(u, i);
      if (d1 > d2) {
        if (res < d1) {
          res = d1;
          u = i;
        }
      } else {
        if (res < d2) {
          res = d2;
          v = i;
        }
      }
    }
    cout << res << ' ';
  }
}
