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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include	<bits/stdc++.h>
using	namespace	std;
typedef	long long ll;

// int/long:		-2,147,483,648 - 2,147,483,647 (-2^31 <= int < 2^31)
// long/long long:	-9,223,372,036,854,775,808 - 9,223,372,036,854,775,807 (-2^63 <= long < 2^63)

// lower_bound(A.begin(), A.end(), N)
// upper_bound(...

// A.erase(unique(A.begin(), A.end()), A.end())

// bit: &/and, |/or, ^/xor, ~/not

// getline(cin, String)
// while (getline(cin, S)) {}

// #include <cstdio>
// int a;
// scanf("%d", &a);
// printf("%d\n", a);
//
// for (int i=0; i<n; ++i)
// 	scanf("%d", &a[i]);
//
// char c;
// scanf(" %c", &c);
//
// cout << "\n";

// cout <<fixed <<setprecision(10)

// priority_queue<int> q;
// priority_queue<int, vector<int>, less<int>> q; // Default: vector, less
// priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> q;

// Graph
// Warshall-Floyd: Distance between each node, N^3
// Dijkstra: Distance from the start node, N^2
// DFS(Depth-First Search)

//
// Notes
//
// When specified less than N, try full scan.
// RE: empty vector
// Check Qs
// long/int
// xor -> count each bit
// Check values
// 	1,000,000 -> Loop
// 	100,000	-> vector(int:400KB, long:800KB) 

#define	INF (1<<30)
//1,073,741,824
//= 536,870,912 *2

#define	MOD 1000000007

/*
ll	myPow(ll x, ll n) {
	if (0==n)
		return 1;
	if (0==n%2)
		return myPow(x*x%MOD, n/2);
	else
		return x*myPow(x, n-1)%MOD;
}
*/
ll	myPow(ll x, ll n) {
	ll r=1;
	while(n>0)
		if (0==n%2) {
			x *=x;
			x %=MOD;
			n /=2;
		} else {
			r *=x;
			r %=MOD;
			n--;
		}
	return r;
}

#define	Rep0(i, n)	for (ll i=0; i<n; i++)
#define	Rep1(i, n)	for (ll i=1; i<=n; i++)
#define Sort(P)	sort(P.begin(), P.end())
#define Rev(P)	reverse(P.begin(), P.end())

int	main() {
	ll N;
	cin >>N;
	vector<ll> A(N, 0);
	vector<ll> S(N+1, 0);
	Rep0(i, N) {
		cin >>A.at(i);
		S.at(A.at(i))++;
	}
	ll SS=0;
	Rep1(i, N)
		SS+=(S.at(i)*(S.at(i)-1))/2;
	Rep0(i, N) {
		ll I=A.at(i);
		ll NN=S.at(I);
		ll SSS=SS-(NN*(NN-1))/2;
		if (NN>2) {
			NN--;
			SSS+=(NN*(NN-1))/2;
		}
		cout <<SSS <<endl;
	}
	return 0;
}