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>
using namespace std;

#define name ""
#define ini freopen(name".inp","r",stdin); freopen(name".out","w",stdout)
#define foe(it,c) for (__typeof(c.begin()) it = c.begin(); it != c.end(); it++)
#define long long long 
#define db double 
#define pii pair <int, int>
#define pll pair <long, long>
#define all(c) c.begin(), c.end()
#define fastio ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

const int INF = 1e9+7;
const int MOD = 1e9+7;
const int dx[4] = {0,0,-1,1};
const int dy[4] = {-1,1,0,0};

int main()
{
  fastio;
  int n; cin >> n;
  vector <int> deg; deg.resize(n + 1, 0);
  for (int i = 1; i < n; i++){
    int u; cin >> u; 
    deg[u]++;
  }
  for (int i = 1; i <= n; i++) cout << deg[i] << '\n';
  return 0;
}