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)

#pragma GCC optimize("Ofast,O3,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;

#define endl '\n'
#define hoise cerr<<"hoise - "<<__LINE__<<"\n";
#define tham getchar();
#define FASTIO ios::sync_with_stdio(false);cin.tie(nullptr);

using ll = long long;
using ii = pair <int, int>;
using vi = vector <int>;
using vl = vector <ll>;

const int MXN = (int)2e5+6;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;

#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}

inline int Set(int N, int pos){return N=N | (1<<pos);}
inline int Reset(int N, int pos){return N=N & ~(1<<pos);}
inline bool Check(ll N, ll pos){return (bool)(N & (1LL<<pos));}

inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }


template <typename T> using orderedSet = tree<T, null_type, greater<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> using orderedSet1 = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
/*--------------------------------------------------------------------------------------------------------------------------*/

vi adj[MXN];


int main(){
	FASTIO
	
	int n;
	cin >> n;
	int a[n+4];
	for (int i = 2; i <= n; ++i) {
		cin >> a[i];
	}
	
	for (int i = 2; i <= n; ++i) {
		adj[i].push_back (a[i]);
		adj[a[i]].push_back (i);
	}
	
	for (int i = 1; i <= n; ++i) {
		cout << (i == 1 ? adj[i].size() : adj[i].size() - 1) << endl;
	}
    return 0;
}

/*

*/