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;
using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using vec_b = vector<bool>;
using vec_str = vector<string>;
using graph = mat;
using edge = pair<ll,ll>;
using wgraph = vector<vector<edge>>;

#define INF (1LL << 60)
#define MOD 1000000007

#define REP(i,m,n) for(ll (i)=(m),i##_len=(n);(i)<(i##_len);++(i))
#define FORR(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define _PR(x) cout << (x) << endl
#define _PS(x) cout << (x) << " "
#define SZ(x) ((ll)(x).size())
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define pb push_back


int main()
{
    ll N;
    cin >> N;
    ll a;
    multiset<ll> S;
    REP(i,0,N-1) {
        cin >> a;
        S.insert(a);
    }
    REP(i,1,N+1) {
        _PR(S.count(i));
    }



    

    return 0;
}

/*



*/