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>

using namespace std;
#define INF 1000007
#define MOD 1000000007
#define maxn 101
#define maxh 1001
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) FOR(i, 0, n)
const int dx[4] = {0, 1, -1, 0}, dy[4] = {1, 0, 0, -1};
typedef std::pair<int, int> P;

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *) array, (T *) (array + N), val);
}

vector<bool> prime(maxn + 1, true); //素数テーブル
void prime_table(void) {
    prime[0] = false;
    prime[1] = false;
    for (int i = 2; i * i <= maxn; i++) {
        if (!prime[i]) {
            continue;
        }
        for (int j = i + i; j <= maxn; j += j + i) {
            prime[j] = false;
        }
    }
}

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b) { return a / gcd(a, b) * b; }

bool IsPrime(int num) //素数判定
{
    if (num < 2)
        return false;
    else if (num == 2)
        return true;
    else if (num % 2 == 0)
        return false;

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2) {
        if (num % i == 0) {
            return false;
        }
    }

    return true;
}

vector<int> divisor(int n) //約数列挙
{
    vector<int> ret;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i - 1);
            if (i * i != n) {
                ret.push_back((n / i) - 1);
            }
        }
    }
    sort(ret.begin(), ret.end());
    return (ret);
}

int a, b;

void f(int x) {
    while (x) {
        a++;
        b = x;
        x /= 10;
    }
}

signed main() {
    int N;
    cin >> N;
    vector<int> A(N + 1);
    rep(i, N - 1) {
        int a;
        cin >> a;
        A[a - 1]++;
    }
    rep(i, N + 1) {
        cout << A[i] << endl;
    }

}