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 M = 1e6 + 3;
long long fac[800005];
long long mpe(long long base, long long exponent, long long modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = result * base % modulus;
    exponent = exponent >> 1;
    base = base * base % modulus;
  }
  return result;
}
int par[200005][18], ht[200005];
long long fi = 1, se = 1;
int lca(int u, int v) {
  if (u == v) return u;
  if (ht[u] < ht[v]) swap(u, v);
  int i = 18;
  for (i = 17; i >= 0; i--)
    if (par[u][i] && ht[par[u][i]] >= ht[v]) {
      u = par[u][i];
    }
  if (u == v) return u;
  i = 17;
  for (i = 17; i >= 0; i--)
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  return par[v][0];
}
long long dis(long long a, long long b) {
  return ht[a] + ht[b] - 2 * ht[lca(a, b)];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  par[1][0] = 0;
  ht[1] = 1;
  long long n, an = 0;
  cin >> n;
  for (long long i = 2; i < n + 1; ++i) {
    long long a;
    cin >> a;
    ht[i] = ht[a] + 1;
    par[i][0] = a;
    for (long long j = 1; j < 18; ++j) par[i][j] = par[par[i][j - 1]][j - 1];
    long long d1 = dis(i, fi);
    long long d2 = dis(i, se);
    if (d1 < d2) {
      swap(d1, d2);
      swap(fi, se);
    }
    if (an < d1) {
      an = d1;
      se = i;
    }
    cout << an << " ";
  }
  return 0;
}
