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)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define all(t) t.begin(), t.end()
#define inrange(i, a, b) (((i)>= min((a), (b))) && ((i) <= max((a), (b))))
typedef vector<ll> vi;
#define fi first
#define se second
const ll inf = 9e18;
const ll mod = 1e9 + 7;

void print(ll a[], ll n){for(ll i=0;i<n;i++){cout<<a[i]<<" ";}cout<<"\n";}
ll power(ll x, ll y, ll M = inf){
    ll ans = 1;
    x %= M;
    while(y){
        if(y&1)
            ans = (x * ans) % M;

        x = (x * x) % M;
        y >>= 1;
    }
    return ans;
}
ll modInverse(ll n) {return power(n, mod-2, mod);}
inline ll mul(ll a, ll b){ return (a * b) % mod; }
inline ll sub(ll a, ll b){ ll c = a - b; if(c < 0) c += mod; return c; }
inline ll add(ll a, ll b){ ll c = a + b; if(c > mod) c -= mod; return c; }
inline ll divi(ll a, ll b){ return mul(a, modInverse(b)); }

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

	ll n;
	cin>>n;
	ll a[n] = {0};
	map<ll, ll> mp;
	for(int i=1;i<n;i++){
		cin>>a[i];
		mp[a[i]]++;
	}
	for(int i=1;i<=n;i++){
		cout<<mp[i]<<"\n";
	}
	
	return 0;
}