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: Wrong Answer)

/*
 * zyrch - Rupanshu Yadav
 */
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <fstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <utility>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <cassert>
#include <sstream>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef double db;
typedef string str;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<pi> vpi;
typedef vector<pl> vpl;

#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL) 

#define forn(i,a,b) for (int i = (a); i < ll(b); ++i)
#define fo(i,a) forn(i,0,ll(a))
#define rforn(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define ro(i,a) rforn(i,0,a)
#define trav(a,x) for (auto& a: x)

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define rsz resize
#define ins insert

#define mp make_pair
#define pb push_back
#define eb emplace_back
#define f first
#define s second

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

const ll INF = 1e18;

ll mod = 1000000007;
inline ll fast_expo(ll base,ll power,ll modulo=mod){
    base%=modulo;
    if (base<0) base+=modulo;
    ll x=base,cnt=power,ans=1;
    while(cnt){
        if (cnt&1) ans=(ans*x)%modulo;
        x=(x*x)%modulo;
        cnt>>=1;
    }
    return ans;
}

ll takemod(ll a, ll mo){
	a %= mo;
	if(a < 0) a+=mo;
	return a;
}

// if mod is prime
ll modinv(ll a, ll mo){
	return takemod(fast_expo(takemod(a,mo), mo-2, mo), mo);
}

//=========================CODE STARTS HERE=============================//

int main() {
	
	ll n;
	cin >> n;
	
	vl arr(n);
	map<int, int> mp;
	fo(i, n) {
		cin >> arr[i];
		mp[arr[i]]+=1;
	}
	
	ll sum = 0;
	map<int, int> diff;
	
	trav(i, mp) {
		sum += (i.s * (i.s - 1))/2;
		if (i.s == 1) diff[i.f] = 0;
		else 
		diff[i.f] = (i.s * (i.s - 1))/2 - ((i.s - 1) * (i.s - 2))/2;
	}
	fo(i, n) {
		cout << sum - diff[arr[i]] << "\n";
	}	
	
	return 0; 
	
}