Nikita plays a new computer game. There are m levels in this game. In the beginning of each level a new class appears in the game; this class is a child-class of the class yi (and yi is called parent-class for this new class). Thus, the classes form a tree. Initially there is only one class with index 1.

Changing the class to its neighbour (child-class or parent-class) in the tree costs 1 coin. You can not change the class back. The cost of changing the class a to the class b is equal to the total cost of class changes on the path from a to b in the class tree.

Suppose that at i -th level the maximum cost of changing one class to another is x. For each level output the number of classes such that for each of these classes there exists some other class y, and the distance from this class to y is exactly x.

Input

First line contains one integer number m — number of queries (1 ≤ m ≤ 3·105).

Next m lines contain description of queries. i -th line (1 ≤ i ≤ m) describes the i -th level and contains an integer yi — the index of the parent-class of class with index i + 1 (1 ≤ yi ≤ i). 

Output

Suppose that at i -th level the maximum cost of changing one class to another is x. For each level output the number of classes such that for each of these classes there exists some other class y, and the distance from this class to y is exactly x.

Examples

Input

4
1
1
2
1


Output

2
2
2
3


Input

4
1
1
2
3


Output

2
2
2
2
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % MOD;
}
template <typename T>
inline bool IsPri(T x) {
  if (x < 2) return false;
  for (T i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
template <typename T>
inline T Gcd(T a, T b) {
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <typename T>
inline int BitCnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
template <typename T>
inline T MDis(T x1, T y1, T x2, T y2) {
  return abs(x1 - x2) + abs(y1 - y2);
}
template <typename T>
inline T MDis(T x1, T y1, T z1, T x2, T y2, T z2) {
  return abs(x1 - x2) + abs(y1 - y2) + abs(z1 - z2);
}
inline double EDis(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
inline double EDis(double x1, double y1, double z1, double x2, double y2,
                   double z2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) +
              (z1 - z2) * (z1 - z2));
}
inline long long Pow(long long a, long long n) {
  long long t = 1;
  a %= MOD;
  while (n > 0) {
    if (n & 1) t = t * a % MOD;
    a = a * a % MOD, n >>= 1;
  }
  return t % MOD;
}
inline int read() {
  static char buf[1000000], *p1 = buf, *p2 = buf;
  register int x = false;
  register char ch =
      p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
          ? EOF
          : *p1++;
  ;
  register bool sgn = false;
  while (ch != '-' && (ch < '0' || ch > '9'))
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  if (ch == '-')
    sgn = true,
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48),
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  return sgn ? -x : x;
}
const int N = 3e5 + 100;
vector<int> v[2];
int fz[N][20], dep[N];
void AddVex(int u, int f) {
  fz[u][0] = f, dep[u] = dep[f] + 1;
  for (int i = 1; i < 20; ++i) fz[u][i] = fz[fz[u][i - 1]][i - 1];
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 19; dep[u] != dep[v]; --i)
    if (dep[fz[u][i]] >= dep[v]) u = fz[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; --i)
    if (fz[u][i] != fz[v][i]) u = fz[u][i], v = fz[v][i];
  return fz[u][0];
}
int Dis(int u, const vector<int> &v) {
  if (!((int)(v).size())) return 0;
  return dep[u] + dep[v[0]] - 2 * dep[LCA(u, v[0])];
}
int main() {
  int n, f, D = 0;
  cin >> n;
  v[0].push_back(1);
  for (int i = 2; i <= n + 1; ++i) {
    scanf("%d", &f), AddVex(i, f);
    int d[2] = {Dis(i, v[0]), Dis(i, v[1])}, mx = max(d[0], d[1]);
    if (mx > D) {
      for (int j = 0; j < 2; ++j)
        if (Dis(i, v[j]) > D) {
          for (int k : v[!j])
            if (Dis(k, {i}) == mx) v[j].push_back(k);
          v[!j].clear();
          v[!j].push_back(i);
          break;
        }
      D = mx;
    } else if (mx == D) {
      for (int j = 0; j < 2; ++j)
        if (Dis(i, v[j]) == D) {
          v[!j].push_back(i);
          break;
        }
    }
    printf("%d\n", ((int)(v[0]).size()) + ((int)(v[1]).size()));
  }
  return 0;
}
