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<limits>
#define ll long long
#define F first
#define S second
#define pb push_back
#define oo (1LL<<63-1)
#define endl '\n'
#define si size()
#define all(v) v.begin(),v.end()
#define FASTIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define Matrix vector< vector < long long > >

using namespace std;

int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};

const long double Pi=acos(-1),e=2.718;
const int N=2e5+10, mod=1e9+7;

int main ()
{
    FASTIO
    int n;
    cin >> n;
    map<int,int> Map;
    for(int i=1;i<=n;i++)Map[i]=0;
    for(int i=2;i<=n;i++){
        ll a;
        cin >> a;
        Map[a]++;
    }
    for(auto x:Map){
        cout << x.S << endl;
    }
    return 0;
}
/**

*/