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 <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include <unistd.h>
#include <math.h>
long long int ans=0;
long long sort(long long d[],int NumberArrays,int up){
  int tmp;
  for (int i = 0; i < NumberArrays-1; i++) {
    if (up && d[i]<d[i+1]) {
      d[i]=tmp;
      d[i]=d[i+1];
      d[i+1]=tmp;
    }else if (d[i]>d[i+1]) {
      d[i]=tmp;
      d[i]=d[i+1];
      d[i+1]=tmp;
    }
  }
}
//最大公約数
long long int getgcd(long long n, long long m)
{
  int r;
  if (n > m) {
    long long t = m;
    m = n;
    n = t;
  }
  while ((r = m % n) != 0) {
    m = n;
    n = r;
  }
return n;
}

//nの階乗を計算
long long factorial(long long n){
    long long int answer = 1;
while(n > 1){
  answer *= n;
  n--;
}
    return answer;
}

//組み合わせの数nCrを計算
long long calcNumOfCombination(long long  n, long long r){
    return factorial(n) / (factorial(r) * factorial(n-r));
}


int main(int argc, char* argv[]){
	long long int a = 0, b = 0,d=0,e=0,c[20050]={0};
	scanf("%lld",&a);
  long long f[20050];
  for (int i = 1; i < a; i++) {
    scanf("%lld",&f[i]);
    c[f[i]]++;
  }
  for (int i = 1; i < a+1; i++) {
    printf("%lld\n",c[i]);
  }
  return 0;
}