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 <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <deque>
using namespace std;
using P = pair<int,int>;
typedef long long int ll;
#define rep(i, n) for (int(i) = 0; (i) < (n); (i)++)
#define PI 3.14159265359
ll gcd(ll a, ll b) {//最大公約数
    if(a<b){
        ll tmp = a;
        a = b;
        b = tmp; 
    }
    if (b==0) return a;
    else return gcd(b, a%b);
}
ll lcm(ll a, ll b) {//最小公倍数
    return a * b / gcd(a, b);
}

int main() {
    int n; cin >> n;
    vector<int>a(n);
    vector<int>num(n,0);
    for(int i=1; i<n; i++){
        cin >> a[i];
        num[a[i]-1]++;
    }
    rep(i,n){
        cout << num[i] << endl;
    }
    return 0;
}