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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
 public class Main {
 
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
      	int N = scanner.nextInt();//会社に何人いるか
      
      //List <Integer> n = new ArrayList<Integer>();//各社員の部下の数を格納
      List <Integer> a = new ArrayList <Integer>();//各社員の上司番号を格納
      int n[] = new int[N];
      for(int i=0; i<N; i++){
      //n.add(0);//各社員の部下の数の表記を一旦0と置く...[0,0,0,0,0]
        n[i] = 0;
      }
      
      for(int i=0; i<N-1; i++){
         a.add(scanner.nextInt());//各社員の上司番号を入れる...[1,1,2,2]
      }
      
      int b = 0;
      for(int i=0; i<N-1; i++){
        b = a.get(i);//[1,1,2,2]をbに一つずつ代入していく
        //n.set(b-1,n.get(b-1)+1);//n.set(0,1)...0番目のところに1を置き換える
        //n.get(b-1)+1はi=0で始まっているため,nの0番目に+1加算していくことを表している
        n[b-1] +=1;//[b-1]番目にずつ1加算していく
      }
      for(int i=0; i<N; i++){
       System.out.println(n[i]);
      }
    }
 }