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)

/*-- ILSH  --*/
#include "bits/stdc++.h"
using namespace std;
typedef long long int ll;
#define pb push_back 
#define pop pop_back
#define mp make_pair
#define vii vector < int >
#define vll vector < ll > 
#define pii pair < int , int >
#define pll pair < ll , ll >
#define vpi vector < pii >
#define vpl vector < pll >
#define fi first
#define sz size 
#define len length 
#define se second 

const ll mod = 1e9+7;
const ll inf=1e18+1e17;
//const ll mod = 998244353;

ll modInverse(ll a,ll m){ll m0=m;ll y=0,x=1;if(m == 1)return 0;while(a> 1){ll q=a/m;ll t=m;m=a%m,a=t;t=y;y=x-q*y;x=t;}if(x<0)x+=m0;return x;}  
ll powm(ll a,ll b){a=a%mod;ll res=1;while(b){if(b&1)res=(res*a)%mod;a=(a*a)%mod;b>>=1;}return (res%mod);}

const ll N=2e5+6;
const long double pi=3.14159265358979323;

vii v[N];
void solve(){
	int n;
	cin>>n;
	for ( int i=0;i<n-1;i++){
		int aa;
		cin>>aa;
		v[aa].pb( i+2);
	}
	for ( int i=1;i<=n;i++)
		cout<<v[i].sz()<<'\n';

	

}


int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    ll t=1;
//	cin>>t;
    for (int i=1;i<=t;i++){
    //	cout<<"Case #"<<i<<": ";
    	solve();
 	}
}