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: Accepted)

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

class Program
{

	//numC2を計算する関数
	static long Combi(long num){
			if(num < 2){
					return 0;
			}else{
					return num * (num-1) / 2;
			}
	}

	//Main
	static void Main(string[] args)
	{
		//読み込む
		long N = long.Parse(Console.ReadLine());
		Dictionary<long, long> combiTable = new Dictionary<long, long>();
		long[] An = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).ToArray();

		//バケツを用意
		long[] bucket = new long[N];
		for(long i=0; i<N; i++){
				//バケツに追加
				bucket[An[i]-1] ++;
		}

		//全て揃っているときの場合の数を格納する配列
		long[] combiFull = new long[N];
		//一つ足りないときの場合の数を格納する配列
		long[] combiMin = new long[N];

		//上記2つの配列を埋める
		for(long i=0; i<N; i++){
				combiFull[i] = Combi(bucket[i]);
				combiMin[i] = Combi(bucket[i] - 1);
		}

		//合計値を出しておく
		long fullSum = combiFull.Sum();

		//各玉が抜かれたときの、場合の数の合計を、ディクショナリに追加
		//iは抜かれた玉の番号（i+1の整数が書かれた玉が抜かれる）
		for(long i=0; i<N; i++){
				long sum = fullSum - combiFull[i];
				sum += combiMin[i];

				//ディクショナリに追加
				combiTable.Add(i+1, sum);
		}

		//それぞれの玉につき、ループ処理
		for(long i=0; i<N; i++){
				long ball = An[i];
				long answer = combiTable[ball];
				Console.WriteLine(answer);
		}
	}
}