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 <bits/stdc++.h>

using namespace std;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long int
#define vll vector<ll> 
#define vpll vector<pair<ll,ll>>
#define tc ll t=0;cin>>t;while(t--)
#define pb push_back
#define pop pop_back
#define all(x) (x).begin(),(x).end()
#define lb lower_bound
#define ub upper_bound
#define endl "\n"
#define max *max_element
#define min *min_element
#define f first
#define s second
#define vli vll::iterator 
#define mp make_pair
#define max_val 99999999
#define min_val -99999999
#define mod 1000000007

int main()
{
    IOS
    #ifdef LOCAL
            freopen("C:\\Users\\Just Coding Things\\Codeforces\\input.txt","r",stdin);
            freopen("C:\\Users\\Just Coding Things\\Codeforces\\output.txt","w",stdout);
        #endif
    ll n,i,cnt;
    cin>>n;
    vll v(n-1),u(n),x;
    for(i=0;i<n-1;i++)
    {
        cin>>cnt;
        u[cnt-1]++;
    }
    
    
    
    
    for(i=0;i<n;i++)
    {
        cout<<u[i]<<"\n";
    }
    cout<<endl;
    return 0;
}