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)

// Author : Sarthak Kapoor
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define rep(i,n) for(int i=0;i<n;++i)
#define repa(i,a,n) for(int i=a;i<n;++i)
#define repr(i,n) for(int i=n-1;i>=0;--i)
#define repba(i,b,a) for(int i=b;i>=a;--i)
#define repab(i,a,b) for(int i=a;i<b;++i)
#define ll long long
#define pi pair<int,int> 
#define pii pair<pi,int>
#define ull unsigned long long
#define vi vector<int>
#define viip vector<pair<int,pair<int,int> > > 
#define mp make_pair
#define vip vector<pair<int,int> >
#define pb push_back 
#define fi first
#define sec second
#define all(v) v.begin(),v.end()
#define s(v) v.size()
ll mod = 1000000007;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());//use rng()%n for numbers in range [0,n-1]
ll gcd(ll a,ll b){
	if(b==0)return a;
	return gcd(b,a%b);
}
ll fastexp(ll x,ll a){
    ll res = 1;
    while(a>0){
        if(a&1){
            res = (res*x)%mod;
        }
        a=a>>1;
        x=(x*x)%mod;
    }
    return res;
}
ll inverse(ll n){
    return fastexp(n,mod-2);
}
template <typename T>
void add(T &a, T b){
	a += b;
	if(a >= mod)a -= mod;
}
template <typename T>
void sub(T &a, T b){
	a -= b;
	if(a < 0)a += mod;
}
template <typename T>
void mul(T &a, T b){
	a *= b;
	if(a >= mod)a %= mod;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin>>n;
	vi v(n);
	map<int,int> mp;
	rep(i,n-1){
		cin>>v[i];
		mp[v[i]]++;
	}	
	repa(i,1,n+1){
		cout<<mp[i]<<"\n";
	}
	return 0;
}