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# (Mono-mcs 6.8.0.105) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace C
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());
            var bossInput = Console.ReadLine().Split(' ');
            var bossNum = new int[N - 1];
            for (int i = 0; i < N - 1; i++) //直属の上司の番号を格納している
            {
                bossNum[i] = int.Parse(bossInput[i]);
            }
            //var ansArray = new int[N]; //部下の人数
            //for (int i = 0; i < N; i++) //1番番号の若いボスから始める
            //{
            //    for (int j = N - 1; j > i; j--) //1番番号の大きい下っ端から始め、今検査してるボスまで続ける
            //    {
            //        if (bossNum[j] - 1 == i)
            //        {
            //            ansArray[i]++;
            //        }
            //    }
            //}

            //for (int i = 0; i < N; i++)
            //{
            //    Console.WriteLine(ansArray[i]);
            //}
            var ans = 0;
            for (int i = 1; i <= N; i++)
            {
                for (int j = N; j > i; j--)
                {
                    if (bossNum[j - 2] == i)
                    {
                        ans++;
                    }
                }
                Console.WriteLine(ans);
                ans = 0;
            }
        }
    }
}