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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define REP(i,n) for(ll i=0; i<(ll)(n); i++)
#define REPM(i,n) for(ll i=(n-1); 0=<i; i--)
#define VI vector<int>
#define VL vector<ll>
#define VD vector<double>
#define VS vector<string>
#define V_SORT_U(x) sort(x.begin(), x.end());
#define V_SORT_D(x) sort(x.begin(), x.end(),greater<int>());
#define V_SUM(x) accumulate(x.begin(), x.end(), 0)
#define STR_TRANS_UP(x) transform(x.begin(), x.end(), x.begin(), ::toupper);
#define STR_TRANS_LOW(x) transform(x.begin(), x.end(), x.begin(), ::tolower);

int main(void){
    ll N;
    cin >> N;

    VL A_i(N+1);
    
    REP(i,N-1)
    {
        cin >> A_i.at(i+1);
    }

    A_i.at(0) = 1;

    for(ll i=1; i<N+1; i++)
    {
        int count = 0;
        for(ll j=i; j<N; j++)
        {
            if(A_i.at(j) == i) count++;
        }

        cout << count << endl;
    }

    return 0;
}