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 <iostream>
#include <vector>
#include <map>
using namespace std;
int main(int argc, char* argv[])
{
	int N,i;
	cin>>N;
	vector<int> A(N);
	for (i=1;i<N;i++){
		cin>>A[i];
	}
	map<int,int> myMap;
	map<int,int>::iterator mit;

	for (i=1;i<N;i++){
		mit=myMap.find(A[i]);
		if (mit==myMap.end()){
			myMap[A[i]]=1;
		}else{
			(*mit).second++;
		}
	}
	for (i=1;i<=N;i++){
		mit=myMap.find(i);
		if (mit==myMap.end()){
			cout<<0<<endl;
		}else{
			cout<<(*mit).second<<endl;
		}
	}
	return 0;
}