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: Wrong Answer)

#include <bits/stdc++.h>
#include <numeric>
#define ll long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)


using namespace std;

const ll mod = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;


#define DEBUG

#ifdef DEBUG
#define PRINT(A) std::cout << (#A) << ":" << (A) << std::endl;
#else
#define PRINT(A)
#endif

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


int A, B;
int a[10010][10010], b[1000];
int dp[10010][3];
 
int main()
{
    ll N; cin >> N;
    ll A[2*100000+1];
    rep(i,N){
        A[i] = 0;
    }
    for(ll i = 2; i <= N; i++){
        ll B;
        cin >> B;
        A[B] += 1;
    }

    for(int i = 1; i <= N; i++){
        cout << A[i] << endl;
    }
    return 0;
}