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>
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair <int, int>
#define pll pair <ll, ll>
#define pci pair <char, int>
#define pld pair <ld, ld>
#define ppld pair <pld, pld>
#define ppll pair <pll, pll>
#define pldl pair <ld, ll>
#define vll vector <ll>
#define vvll vector <vll>
#define vpll vector <pll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define mll map <ll, ll>
#define fastmap gp_hash_table
#define cd complex <double>
#define vcd vector <cd>
#define PI 3.14159265358979
#define ordered_set tree <ll, null_type, less <ll>, rb_tree_tag, tree_order_statistics_node_update>
#pragma 03
using namespace std;
using namespace __gnu_pbds;
ll cnt[200005];
int main(){
	ll n; cin >> n;
	for (ll i = 0; i < n - 1; i++){
		ll ip; cin >> ip; cnt[ip]++;
	}
	for (ll i = 1; i <= n; i++) cout << cnt[i] << endl;
}