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>
#include <math.h>
#include <iomanip>      // std::setprecision
using namespace std;
#define what_is(x) cerr << #x << " is " << x << endl;
#define bit(x,i) (x&(1<<i))  //select the bit of position i of x
#define lowbit(x) ((x)&((x)^((x)-1))) //get the lowest bit of x
#define hBit(msb,n) asm("bsrl %1,%0" : "=r"(msb) : "r"(n)) //get the highest bit of x, maybe the fastest
#define IN(i,l,r) (l<i&&i<r) //the next for are for checking bound
#define LINR(i,l,r) (l<=i&&i<=r)
#define LIN(i,l,r) (l<=i&&i<r)
#define INR(i,l,r) (l<i&&i<=r)
#define F(i,L,R) for (int i = L; i < R; i++) //next four are for "for loops"
#define FLL(i,L,R) for (long long int i = L; i < R; i++) //next four are for "for loops"
#define FE(i,L,R) for (int i = L; i <= R; i++)
#define FF(i,L,R) for (int i = L; i > R; i--)
#define FFE(i,L,R) for (int i = L; i >= R; i--)
#define char2Int(c) (c-'0')
#define lastEle(vec) vec[vec.size()-1] 
#define SZ(x) ((int)((x).size()))
#define REMAX(a,b) (a)=max((a),(b)) // set a to the maximum of a and b
#define REMIN(a,b) (a)=min((a),(b));
#define FENWICK(k) (k&-k)
#define FOREACH(i,t) for (auto i=t.begin(); i!=t.end(); i++) // traverse an STL data structure
#define ALL(c) (c).begin(),(c).end() //handy for function like "sort()"
#define PRESENT(c,x) ((c).find(x) != (c).end()) 
#define CPRESENT(c,x) (find(ALL(c),x) != (c).end()) 
#define LL long long //data types used often, but you don't want to type them time by time
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define IOS ios_base::sync_with_stdio(0); //to synchronize the input of cin and scanf
#define INF 1001001001
#define PI 3.1415926535897932384626
//for map, pair
#define mp make_pair
#define fi first
#define se second
//for vectors
#define pb push_back
typedef int elem_t;
typedef vector<int> vi; 
typedef vector<vi> vvi; 
typedef pair<int,int> ii; 
// directions
const int fx[4][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}};
const int fxx[8][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}, {1,1}, {1,-1}, {-1,1}, {-1,-1}};
const int rogi[4][2] = { {1,1}, {1,-1}, {-1,1}, {-1,-1} };

int n;
int ile[1000007];

void dane() {
    cin>>n;
    F(i,0,n-1) {
        int a;
        cin>>a;
        ile[a]++;
    }
    F(i,1,n+1) cout<<ile[i]<<endl;
}

int main() {
	IOS
	// int t;
	// cin>>t;
	// F(i,0,t)
	    dane();
    return 0;
}