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 ll long long int
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define mem(x,y) memset(x,y,sizeof(x))//memset only use to initialize 0 or -1 no other number.
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mod 1000000007
//#define mod 998244353
using namespace std;
using namespace __gnu_pbds; 
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> os;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> os_pair;

ll pow(ll x,ll n){ll res =1;while(n>0){if(n%2==1){res=res*x;}x=x*x;n=n/2;}return res;}
ll powm(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modinv(ll a){
	return powm(a,mod-2)%mod;
}
int main(){

ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    t=1;
    //cin>>t;
    while(t--){
    	int n,i,j;
    	
    	cin>>n;
    	int a[n+1];
    	mem(a,0);
    	
    	for(i=0;i<n-1;i++)
    	{
    		int x;
    		cin>>x;
    		a[x]++;
    		
		}
		for(i=1;i<n+1;i++)
		{
			cout<<a[i]<<"\n";
		}
    	
    }}