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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include<math.h>
#include<sstream>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld; 
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ll> vll;




#define F first
#define S second
#define pob pop_back
#define pb push_back
#define Time()	cerr << clock() * 1000 / CLOCKS_PER_SEC << "ms\n"
#define show(x)	cerr << #x << " = " << (x) << '\n'
#define FASTER ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define invec(v) for(auto& x : v)cin>>x;
#define Test int T;cin>>T;while(T--)
#define REP(i,j,k) for(int i=(j);(i)<(k);(i)++)
//#define phi ((1+sqrt(5))/2)
#define endl '\n'
const int MxN = 1e9+7;
const double pi = 2*acos(0.0);
int main(){
	//cout.precision(15);
	FASTER;
	unordered_map<int,int> m;
	int n;
	cin>>n;
	vi v(n-1);
	//invec(v);
	for(int& x: v){
		cin>>x;
		m[x]++;
	}
	for(int i=0;i<n;i++){
		cout<<m[i+1]<<endl;
	}
	
}