Problem:
Score : 300 points

Problem Statement

A company has N members, who are assigned ID numbers 1, ..., N.

Every member, except the member numbered 1, has exactly one immediate boss with a smaller ID number.

When a person X is the immediate boss of a person Y, the person Y is said to be an immediate subordinate of the person X.

You are given the information that the immediate boss of the member numbered i is the member numbered A_i. For each member, find how many immediate subordinates it has.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i < i

Input

Input is given from Standard Input in the following format:

N
A_2 ... A_N

Output

For each of the members numbered 1, 2, ..., N, print the number of immediate subordinates it has, in its own line.

Sample Input 1

5
1 1 2 2

Sample Output 1

2
2
0
0
0

The member numbered 1 has two immediate subordinates: the members numbered 2 and 3.

The member numbered 2 has two immediate subordinates: the members numbered 4 and 5.

The members numbered 3, 4, and 5 do not have immediate subordinates.

Sample Input 2

10
1 1 1 1 1 1 1 1 1

Sample Output 2

9
0
0
0
0
0
0
0
0
0

Sample Input 3

7
1 2 3 4 5 6

Sample Output 3

1
1
1
1
1
1
0

Solution in C++ (Clang 10.0.0) (judge verdict: Accepted)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#define all(x) (x).begin(), (x).end()
using namespace std;
using ll = long long;

template <typename T> inline constexpr bool chmin(T &a, T b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T> inline constexpr bool chmax(T &a, T b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
template <typename ite>
void print_ite(ite start, ite end, char sp = ' ', char en = '\n') {
  for (auto i = start; i != end; i++)
    cout << *i << sp;
  cout << en;
}
template <typename ite> void scan_ite(ite start, ite end) {
  for (auto i = start; i != end; i++)
    cin >> (*i);
}
inline void Yes(bool condition) { cout << (condition ? "Yes\n" : "No\n"); }

int main() {
  int n;
  cin >> n;
  vector<int> cnt(n, 0);
  for (int i = 0; i < n - 1; i++) {
    int a;
    cin >> a;
    a--;
    cnt[a]++;
  }
  for (auto i : cnt) {
    cout << i << endl;
  }
}