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 <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>


#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  ll_cout(X) printf("%lld\n",X)

typedef long long  ll;
#define rept(x, s, n) for (int x = s; x < n; x++)

#define Yes_RTN s_cout("Yes");return 0
#define No_RTN  s_cout("No");return 0
#define YES_RTN s_cout("YES");return 0
#define NO_RTN  s_cout("NO");return 0


int i_cins(int n, int* A);
int l_cins2(int n, long* A,long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);

int s_asorts(const void* a, const void* b);

//int main(void) {
//	int  R;
//	double tmp;
//
//	i_cin(R);
//
//	f_cout((double)R * 3.14*2.0);
//
//	return 0;
//}
static long  a[200000];

int main(void) {
	int N,i;
	int A[200001];

	int cnt = 0;


    i_cin(N);
	i_cins(N-1, A);

	for (i = 0; i < N-1; i++)
		a[A[i]-1]++;
	
	
	for (i = 0; i < N; i++) {
		l_cout(a[i]);
	}
	
	return 0;
}
//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}	
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a -*(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}