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: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.HashMap;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BannedK solver = new BannedK();
        solver.solve(1, in, out);
        out.close();
    }

    static class BannedK {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] k = new int[n];

            HashMap<Integer, Integer> counter = new HashMap<>();
            for (int i = 0; i < n; i++) {
                int a = in.nextInt();
                counter.put(a, counter.getOrDefault(a, 0) + 1);
                k[i] = a;
            }

            long result = 0;
            for (Integer key : counter.keySet()) {
                long anum = counter.get(key);
                result += anum * (anum - 1) / 2;
            }

            for (int i = 0; i < n; i++) {
                int a = k[i];
                int anum = counter.get(a);
                long before = anum * (anum - 1) / 2;
                long after = (anum - 1) * (anum - 2) / 2;
                out.println(result - before + after);
            }
        }

    }
}