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)

/* Simplicity and Goodness */
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update> indexed_set;

void my_dbg() { cout << endl; }
template<typename Arg, typename... Args> void my_dbg(Arg A, Args... B) 
{ cout << ' ' << A; my_dbg(B...); }
#define dbg(...)  cout << "(" << #__VA_ARGS__ << "):", my_dbg(__VA_ARGS__)

#define scn(n) scanf("%d", &n)
#define lscn(n) scanf("%lld", &n)
#define pri(n) printf("%d ", (int)(n))
#define prin(n) printf("%d\n", (int)(n))
#define lpri(n) printf("%lld ", n)
#define lprin(n) printf("%lld\n", n)
#define rep(i,a,b) for(int i=(int)(a); i<(int)(b); i++)
#define pb push_back
#define mp make_pair
#define F first
#define S second

using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;

const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9+7;
const int N = 2e5+2;

void solve()
{
    int n;
    scn(n);

    vi fa(n+1, 0);

    rep(i, 0, n-1) {
        int val;
        scn(val);
        fa[val] ++;
    }

    rep(i, 1, n+1) {
        prin(fa[i]);
    }
}

int main()
{
    int t = 1;
    // scn(t);

    while(t --) {
        solve();
    }
    return 0;
}