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: Wrong Answer)

#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <iomanip>
#include <deque>
#include <set>

typedef long long ll;
#define rep(i,a,b) for(int i=a;i<b;++i)
#define rrep(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
using vi = vector<int>;
using vll = vector<ll>;
ll mod = 1e9 + 7;
//ll llmax = 10000000000000000000ll;
using namespace std;
using Graph = vector<vector<int>>;
Graph G;
const int AMAX = 100000;
/*int cnt_digit(ll N)
{
	int digit = 0;

	while (N > 0)
	{
		N /= 10;
		digit++;
	}
	return digit;
}



int n;
vll a;

ll solve(bool isp)
{
	ll sum = 0ll;
	ll ret = 0ll;

	for (int i = 0; i < n; i++)
	{
		sum += a[i];
		if (isp and sum <= 0)
		{
			ret += -sum + 1;
			sum = 1ll;
		}
		if (not isp and sum >= 0)
		{
			ret += sum + 1;
			sum = -1ll;
		}
		isp ^= 1;
	}
	return ret;
}

*/


/*
 * 最大公約数計算
 */
ll gcd(ll a, ll b)
{
	if (b == 0) {
		return a;
	}
	else {
		return gcd(b, a % b);
	}
}

ll lcm(ll a, ll b)
{
	ll g = gcd(a, b);

	return a / g * b;
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	int N;
	cin >> N;
	vector<int> mem(N - 1);
	map <int,int> data;
	for (auto& i : mem)
	{
		cin >> i;
	}
	for (auto& i : mem)
	{
		data[i] = 0;
	}
	for (auto& i : mem)
	{
		data[i]++;
	}
	for (int i = 1;i<N;i++)
	{
		cout << data[i] << endl;
	}
	
	
	return 0;

}