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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

public class MainClass
{
	public const long Giri = 1000000007;
	public const long DpInf = 99999999999999;
	
	
	public static void Main(string[] args)
	{
 		var n = Console.ReadLine().ToInt();
		var ai = Input();
		var a = new long[n];
		for(int i = 0; i < n; i++)
		{
			a[i] = ai[i].ToLong()-1;
		}
		var map = new long[n];
		long max = 0;
		for(int i = 0; i < n; i++)
		{
			map[a[i]]++;
			max = Math.Max(max, a[i]);
		}
		max++;
		
		var cs = new long[max];
		long all = 0;
		for(int i = 0; i < max; i++)
		{
			cs[i] = Combination(map[i], 2);
			all += cs[i];
		}
		
		for(int i = 0; i < n; i++)
		{
			long c = 0;
			
			if(cs[a[i]] > 1)
			{
				c = Combination(map[a[i]]-1, 2);
			}
			Console.WriteLine(all - cs[a[i]] + c);
			
		}
		
	}
	
	static long ModPow(long a, long n, long mod)
	{
	    long res = 1;
	    while (n > 0)
	    {
	        if ((n & 1) == 1) res = res * a % mod;
	        a = a * a % mod;
	        n >>= 1;
	    }
	    return res;
	}
	
	//コンビネーション 繰り返し二乗法使うやつ nCk
	static long Combination(long n, long k)
	{
		long q = 1;
		long s = 1;
		long j = 1;
		for(long i = n; i >= n-k+1; i--)
		{
			q *= i % Giri;
			q %= Giri;
			s *= ModPow(j, Giri - 2, Giri) % Giri;
			s %= Giri;
			j++;
		}
		return (long)(q * s) % Giri;
	}
	
	static long GCD(long a, long b)
	{
		var v = new[] { a, b };
		while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
		return v[0];
	}
	public static string[] Input() { return Console.ReadLine().Split(); }
}

public static class Smarter
{
	public static int ToInt(this string str) { return int.Parse(str); }
	public static long ToLong(this string str) { return long.Parse(str); }
	public static int[] ToInt(this string[] str, int offset = 0)
	{
		var r = new int[str.Length];
		for(int i = 0; i < str.Length; i++) r[i] = str[i].ToInt() + offset;
		return r;
	}
}