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 <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
 
using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
#define int ll
#define pb push_back
#define pii pair<int,int>
#define ff first
#define ss second
const int MAX = 2e5+10;
typedef tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
int par[MAX];



signed main(){

	ios_base::sync_with_stdio(false);
	cin.tie(0);


	


	int n; cin>>n;
	for (int i=0;i<n-1;i++){
		int x; cin>>x;
		par[x]++;
	}
	
	for(int i=1;i<=n;i++) cout<<par[i]<<"\n";
	return 0;
}