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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

namespace abc163c
{
    class Program
    {
        static void Main(string[] args)
        {
            //==========================================================================================================================
            //1) input
            int n = int.Parse(Console.ReadLine());
            List<int> a = Console.ReadLine().Split(' ').Select(int.Parse).ToList();

            //==========================================================================================================================
            //2) output
            Dictionary<int, int> holder = new Dictionary<int, int>();
 
            for (int index = 0; index < n - 1; index++)
            {
                if (holder.ContainsKey(a[index]))
                {
                    holder[a[index]]++;
                }
                else
                {
                    holder.Add(a[index], 1);
                }
            }


            for (int index = 1; index <= n; index++)
            {
                int count = 0;

                if (holder.ContainsKey(index))
                {
                    count = holder[index];
                }

                Console.WriteLine(count);

            }


        }
    }
}