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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)


import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] A = new int[N];
		for(int i =0;  i<N; i++){
			A[i] = sc.nextInt();
		}
		int[] count = new int[N];

		for(int a : A) {
			count[a-1] ++;
		}

		long[] choice2 = new long[N+1];
		combination.intFactorial();
		for(int i=0; i<N+1; i++) {
			choice2[i] = combination.comb(i, 2);
			//System.out.println(choice2[i]);
		}

		for(int k=0; k < N; k++) {
			long sum = 0;
			for(int i =0;  i<N; i++) {
				if(i == A[k]-1) {
					//System.out.print(count[i]-1 + " ");
					try{
						sum += choice2[count[i]-1];
					}catch(Exception e) {
					}
				}else {
					//System.out.print(count[i] + " ");
					try{
						sum += choice2[count[i]];
					}catch(Exception e) {
					}
				}
			}
			System.out.println(sum);
		}




	}
}

class combination{


	static final int MOD = 1000000007;

	//階乗時の各値をMODの元で計算しておく
	static long[] factorial = new long[10000000];
	static void intFactorial() {
		factorial[0] = 1;
		for(int i = 1; i < 10000000; i++) {
			factorial[i] = factorial[i-1] * i % MOD;
		}

	}

	//フェルマーの小定理を利用して,pを法としてのaの逆元a^p-2を求める
	public static long MODinv(long a ,long p) {
		return modpow(a ,p-2 ,p);
	}

	//a^n (mod p)を二分累乗法で計算する
	public static long modpow(long a, long n, long p) {
		long ans = 1;
		while(n > 0) {
			if(n%2 == 1) {
				ans = ans * a % p;
			}
			n = n / 2;
			a = a * a % p;
		}
		return ans;
	}

	//nCrをMODを法として計算する
	static long comb(int n, int r) {
		if(n==0 && r==0) return 1;
		if(n<r || n<0) return 0;

		return (factorial[n] * MODinv(factorial[r], MOD)) % MOD * MODinv(factorial[n-r], MOD) % MOD;
	}








}