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)

//last updated : 11.04.2020
#include <bits/stdc++.h>
#pragma GCC optimization ("O3")
using namespace std;

#define ll              long long
#define ld              long double
#define pll             pair <ll, ll>
#define plll            pair <ll, pll>

#define fa(i, n, N)     for (ll i = n; i <= N; ++i)
#define fac(i, n, b, N) for (ll i = n; i <= N && b; ++i)
#define fd(i, n, N)     for (ll i = n; i >= N; --i)

#define mp              make_pair
#define pb              push_back
#define pob             pop_back
#define pf              push_front
#define pof             pop_front
#define fi              first
#define se              second

#define ms(arr,t)       memset (arr, t, sizeof(arr));
#define all(x)          x.begin(), x.end()
#define len             length()
#define endl            '\n'

#define isss(x, s)      istringstream x (s)
#define db(x)           cout << ">>>>>> " << #x << " -> " << x << endl
#define AAAA            cout << "\t------" << endl;

//urgent cases only!
#define get_rt		cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms.\n"

const ll inf = INT_MAX;
const ll inf2 = 1e18;
const ll MAX = 1e5+5;
const ll MOD = 1e9+7;
const double pi = acos(-1.0);
const double EPS = 1e-9;

void init(int precision){
  	ios_base::sync_with_stdio(false);
  	cin.tie(0);
  	cout.tie(0);
  	cout.setf(ios::fixed);
  	cout.setf(ios::showpoint);
  	cout.precision(precision);
}

inline void iofile (){
	freopen ("input.txt", "r", stdin);
	// freopen ("output.txt", "w", stdout);
}


ll n, k, arr[MAX*2];

int main() {
    init(10);
    //iofile();
	
	cin >> n;
	fa (i,2,n){
		cin >> k;
		arr[k] ++;
	}
	
	fa (i,1,n){
		cout << arr[i] << endl;
	}

	return 0;
}