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 rep(i,n) for (int i = 0; i < (n); ++i)
#define drep(i,n) for(int i = (n-1); i >= 0; i--)
#define all(v) (v).begin(),(v).end()
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
template <class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template <class T> T lcm(T a, T b) { return a/gcd(a,b)*b; }
typedef pair<int, int> P;
typedef long long ll;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;
const ll MOD = 1e9+7;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
	int n;
	cin >> n;
    vector<int> ans(n);
    int a;
    rep(i,n-1) {
        cin >> a;
        ans[a-1] += 1;
    }
    rep(i,n) {
        cout << ans[i] << endl;
    }
	return 0;
}