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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define ll long long

int gcd(int a, int b)
{
  return b ? gcd(b, a % b) : a;
}

int tgcd(int a, int b, int c)
{
  return gcd(gcd(a, b), c);
}

int main(void)
{
  int n, i, j, k;
  scanf("%d", &n);
  int *a, *b;
  int z;
  a = (int *)malloc(sizeof(int) * n);
  b = (int *)malloc(sizeof(int) * n);
  b[0] = 0;
  for (i = 0; i < n; i++)
  {
    a[i] = 0;
  }
  for (i = 1; i < n; i++)
  {
    scanf("%d", &z);
    b[i] = z;
  }
  for (i = n - 1; i > 0; i--)
  {
    z = b[i];
    a[z - 1]++;
  }

  for (i = 0; i < n; i++)
  {
    printf("%d\n", a[i]);
  }

  return 0;
}