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 vi vector<int>
#define vvi vector<vector<int>>
#define pii pair<int,int>
#define vs vector<string>
#define all(v) (v).begin(),(v).end()
#define ll long long
#define ull unsigned long long
#define fastio ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define ld long double
#define mod 1000000007
#define PI 3.141592653589793238
#define phi 1.638
#define eps 1e-7
//#define endl "\n"
using namespace std;
using namespace __gnu_pbds;
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; 

int main() 
{
    fastio
    int n;
    cin>>n;
    vector<int> a(n);
    for(auto& it : a) cin>>it;

    int ans[n+1] = {0};
    for(int i = 0; i<n; ++i) {
        ans[a[i]]++;
    }

    for(int i = 1; i<=n; ++i) {
        cout<<ans[i]<<endl;
    }
}