Problem:
Score : 400 points

Problem Statement

We have N balls. The i-th ball has an integer A_i written on it.

For each k=1, 2, ..., N, solve the following problem and print the answer.

Find the number of ways to choose two distinct balls (disregarding order) from the N-1 balls other than the k-th ball so that the integers written on them are equal.

Constraints

3 \leq N \leq 2 \times 10^5

1 \leq A_i \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

For each k=1,2,...,N, print a line containing the answer.

Sample Input 1

5
1 1 2 1 2

Sample Output 1

2
2
3
2
3

Consider the case k=1 for example. The numbers written on the remaining balls are 1,2,1,2.

From these balls, there are two ways to choose two distinct balls so that the integers written on them are equal.

Thus, the answer for k=1 is 2.

Sample Input 2

4
1 2 3 4

Sample Output 2

0
0
0
0

No two balls have equal numbers written on them.

Sample Input 3

5
3 3 3 3 3

Sample Output 3

6
6
6
6
6

Any two balls have equal numbers written on them.

Sample Input 4

8
1 2 1 4 2 1 4 1

Sample Output 4

5
7
5
7
7
5
7
5

Solution in Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        HashMap<Integer,Integer> hash = new HashMap<>();
        int[] a = new int[n];
        for (int i=0;i<n;i++){
            a[i] = sc.nextInt();
            if (hash.containsKey(a[i]))hash.put(a[i],hash.get(a[i])+1);
            else hash.put(a[i],1);
        }
        for (int i=0;i<n;i++){
            hash.put(a[i],hash.get(a[i])-1);
            long ans = 0;
            for (int j:hash.values()){
                ans+=combination(j,2);
            }
            hash.put(a[i],hash.get(a[i])+1);
            System.out.println(ans);
        }
    }
    static long factorial(long num){
        if (num<2)return 1;
        else return num*factorial(num-1);
    }
    static int arraySum(int[] array){
        int ans = 0;
        for (int value : array) ans += value;
        return ans;
    }
    static long combination(long a, long b){
        long data = 1;
        for (int i=0;i<b;i++){
            data*=a-i;
        }
        return data/factorial(b);
    }
    static double log(double base,double antilogarithm){
        return Math.log(antilogarithm)/Math.log(base);
    }

    static long gcd(long x,long y){
        if (y==0)return x;
        else return gcd(y,x%y);
    }
    static long lcm(long x,long y){
        return x/gcd(x,y)*y;
    }
    static HashMap<Long,Long> factorization(long num){
        HashMap<Long,Long> hash = new HashMap<>();
        long n = num;
        long count = 2;
        while (n>1){
            while (n%count==0){
                n/=count;
                if (hash.containsKey(count))hash.put(count,hash.get(count)+1);
                else hash.put(count,1L);
            }
            count++;
        }
        return hash;
    }

}