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 = 200000 + 7;
const int M = 59;
const int mid = M / 2;
const int mod = 1e9 + 9;
const int inf = 1e6 + 3;
const long long linf = 1ll * inf * inf;
const double pi = acos(-1);
const double eps = 1e-7;
const double ep = 1e-5;
const int maxn = 1e5 + 7;
const double PI = acos(-1);
int n;
int p[N][20];
int d[N];
set<int> st;
int LCA(int u, int v) {
  int oldU = u, oldV = v;
  if (d[u] < d[v]) swap(u, v);
  while (d[u] > d[v]) {
    int k = log2(d[u] - d[v]);
    u = p[u][k];
  }
  if (u == v) return u;
  for (int j = 19; j >= 0; --j) {
    if (p[u][j] != -1 && p[v][j] != -1 && p[u][j] != p[v][j]) {
      u = p[u][j];
      v = p[v][j];
    }
  }
  return p[u][0];
}
int dist(int u, int v) {
  int x = LCA(u, v);
  return d[u] + d[v] - 2 * d[x];
}
void add(int u, int r) {
  p[u][0] = r;
  d[u] = d[r] + 1;
  for (int j = (1); j < (20); j++) {
    if (p[u][j - 1] == -1)
      p[u][j] = -1;
    else
      p[u][j] = p[p[u][j - 1]][j - 1];
  }
}
int rm[N];
int k = 0;
void solve() {
  int n;
  cin >> n;
  for (int j = (0); j < (20); j++) p[1][j] = -1;
  int px = 1, py = 1;
  int mLen = 0;
  for (int u = (2); u < (n + 1); u++) {
    int v;
    scanf("%d", &v);
    add(u, v);
    int l1 = dist(u, px);
    int l2 = dist(u, py);
    if (l1 > mLen) {
      mLen = l1;
      py = u;
    } else {
      if (l2 > mLen) {
        mLen = l2;
        px = u;
      }
    }
    printf("%d ", mLen);
  }
}
int main() {
  int T = 1;
  for (int i = (1); i < (T + 1); i++) {
    solve();
  }
}
