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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;
use std::collections::HashMap;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn read_vec<T: FromStr>(n: usize) -> Vec<T> {
    return (0..n).map(|_| read()).collect();
}

fn main() {
    let n: usize = read();
    let ai: Vec<usize> = read_vec(n-1);

    let mut map = HashMap::new();
    for a in ai.iter() {
        if map.contains_key(a) {
            if let Some(x) = map.get_mut(a) {
                *x += 1;
            }
        } else {
            map.insert(a, 1);
        }
    }

    for i in 1..n+1 {
        if map.contains_key(&i) {
            println!("{}", map[&i]);
        } else {
            println!("{}", 0);
        }
    }
}