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>
#define  forn(i, n)  for(int i=0; i < (int)n; i++)
#define  vi          vector<int>
// #define  pb          push_back
// #define  ap          append
// #define  mii         map<int, int> 
using namespace std ;
// vector<string> splitstring(string str, string delimiter) {
//     vector<string> v;
//     string token;
//     istringstream parse(str);
//     while(getline(parse, token, delimiter)) {
//     v.push_back(token);
//     }
//     return v;
// }
// int gcd (int a, int b) {
//     if (b == 0) return a;
//     else return gcd (b, a % b);
// }
// int lcm(int a, int b){
//   return a / gcd(a,b) * b;
// }
int a[200010];
int ans[200010];   
void solve() {
   int n ;
   cin >> n ;
   for(int i=2; i<=n; i++) cin >> a[i] ;
   for(int j=2; j<=n; j++) ans[a[j]]++ ;
   for(int i=1; i<=n; i++) cout << ans[i] << "\n";  
}
int main() {
  int t=1;
  //cin >> t ;
  while(t--) solve() ;
  return 0 ;
}