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: Runtime Error)

#include <bits/stdc++.h>

#define ll long long
#define endl '\n'

using namespace std;


// Returns the index of n if it is in the vector, or -1 case don't
ll bsearch(int n, int vector[], int begin, int end) {
    int l = begin, r = end;
    int mid;

    while(l <= r) {
        mid = (l + r) / 2;

        if(vector[mid] > n) {
            r = mid - 1;
        }

        else if(vector[mid] < n) {
            l = mid + 1;
        }

        else {
            return mid;
        }
    }

    return -1;
}

ll bsearch(ll n,vector<ll> &array, ll begin, ll end) {
    int l = begin, r = end;
    int mid;

    while(l <= r) {
        mid = (l + r) / 2;

        if(array[mid] > n) {
            r = mid - 1;
        }

        else if(array[mid] < n) {
            l = mid + 1;
        }

        else {
            return mid;
        }
    }

    return -1;
}


int main() {
    ll n;
    cin >> n;
    vector<ll> v(n);
    ll i;

//    ios::sync_with_stdio(false);

    for(ll i = 1; i < n; i++) {
        cin >> v[i];
    }

    vector<ll> sub(n);
 
    // N == 1;
    for(i = 1; v[i] == 1; i++) {
        sub[0] += 1;
    }

    for(; i < n; i++) {
        sub[bsearch(v[i] - 1, v, 0, i - 1)] += 1;        
    }

    for(i = 0; i < n; i++) {
        cout << sub[i] << endl;
    }

    return 0;
}