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;
int n, x, y, z, xy, xz, yz, L[200100], T[200100], P[200100][20];
vector<int> graph[200100];
void dfs(int u, int lvl) {
  L[u] = lvl;
  for (int i = 0; i < graph[u].size(); i++) {
    int v = graph[u][i];
    dfs(v, lvl + 1);
  }
}
void build() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; (1 << j) < n; j++) {
      P[i][j] = -1;
    }
  }
  for (int i = 0; i < n; i++) P[i][0] = T[i];
  for (int j = 1; (1 << j) < n; j++) {
    for (int i = 0; i < n; i++) {
      if (P[i][j - 1] != -1) P[i][j] = P[P[i][j - 1]][j - 1];
    }
  }
}
int lca(int u, int v) {
  if (L[u] < L[v]) swap(u, v);
  int log;
  for (log = 1; (1 << log) <= L[u]; log++)
    ;
  log--;
  for (int i = log; i >= 0; i--) {
    if (L[u] - (1 << i) >= L[v]) u = P[u][i];
  }
  if (u == v) return u;
  for (int i = log; i >= 0; i--) {
    if (P[u][i] != -1 && P[u][i] != P[v][i]) {
      u = P[u][i];
      v = P[v][i];
    }
  }
  return T[u];
}
int dist(int u, int v) { return L[u] + L[v] - 2 * L[lca(u, v)]; }
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> x;
    x--;
    T[i] = x;
    graph[x].push_back(i);
  }
  dfs(0, 0);
  build();
  x = y = 0;
  for (int i = 1; i < n; i++) {
    xy = dist(x, y);
    xz = dist(x, i);
    yz = dist(y, i);
    if (xz > xy && xz >= yz)
      y = i;
    else if (yz >= xz && yz > xy)
      x = i;
    cout << max(xy, max(xz, yz)) << " ";
  }
  return 0;
}
