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 <math.h>

int main(){
    int N;
    int *ip; //A_i
    int *count;
    int i;
    int j;
    //N入力
    while(scanf("%d",&N)!=1);
    //メモリの確保
    ip =(int *)malloc((N-1)*sizeof(int)); //A_i
    count =(int *)malloc((N)*sizeof(int)); //出力
    //A_i 入力
    for (i=0;i<N-1;i++){
        while(scanf("%d",&ip[i])!=1);
    }
    //出力結果初期化
    for(i=0;i<N;i++){
        count[i]=0;
    }
    //計算処理
    for(i=0;i<N-1;i++){
        count[ip[i]-1]+=1;
    }
    //最終出力 
    for(i=0;i<N;i++){
        printf("%d\n",count[i]);
    }
    free(ip);
    free(count);
    return 0;
}