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)

#include <bits/stdc++.h>
#ifdef HIRO
#define d(f_, ...) printf((f_), ##__VA_ARGS__)
#else
#define d(f_, ...)
#endif

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef queue<int> qi;
typedef queue< pair<int, int> > qp;
#define mp make_pair
#define pi 3.14159265358979323846
#define MAX max_element
#define MIN min_element
#define all(v) v.begin(), v.end()
#define gt(T) greater<T>()
#define sum(v) accumulate(v.begin(), v.end(), 0)
#define uni(v) do{ \
  sort(v.begin(), v.end()); \
  decltype(v)::iterator result = unique(v.begin(), v.end()); \
  v.erase(result, v.end()); \
}while(0)
#define fi(init, n) for(int i = init; i < n; ++i)
#define fj(init, n) for(int j = init; j < n; ++j)
#define fk(init, n) for(int k = init; k < n; ++k)
#define zero(x) memset(x, 0, sizeof(x))
#define negative(x) memset(x, ~0, sizeof(x))
const ll MOD = 1e9 + 7;
const int INF = 1e9;
const int di[] = { 0, 1, 0, -1 };
const int dj[] = { 1, 0, -1, 0 };

template<typename T>
T gcd(T a, T b) {
	return b ? gcd(b, a % b) : a;
}
template<typename T>
T lcm(T a, T b) {
	return a * b / gcd(a, b);
}
template<typename T>
int finder(vector<T> vec, T number) {
	auto itr = find(vec.begin(), vec.end(), number);
	size_t index = distance(vec.begin(), itr);
	if (index != vec.size()) return 1;
	else return 0;
}
ll frac(ll n){
	if (n == 0) return 1;
	return (n * frac(n - 1)) % MOD;
}
template<class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template<class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
ll info[200010][2];
int main(int argc, char** argv) {
#ifdef HIRO
  string quiz(argv[1]);
  string id(argv[2]);
  string filename("input/" + quiz + "/" + id + ".txt");
  ifstream fin(filename);
  if (fin.peek() == ifstream::traits_type::eof()) {
    cout << "No input.";
    return 0;
  }
  ifstream cin(filename);
  std::cin.rdbuf(cin.rdbuf());
#endif
  ll ans = 0;
  ll n;
  cin >> n;
  // ll n, m;
  // cin >> n >> m;
  // ll n, m, k;
  // cin >> n >> m >> k;
  zero(info);
  int arr[n];
  fi(0, n) {
    cin >> arr[i];
    info[arr[i]][0]++;
  }
  ll total = 0;
  fi(0, 20010) {
    if(info[i][0] > 1){
      info[i][1] = info[i][0] * (info[i][0] - 1) / 2;
      total += info[i][1]; 
    }
  }
  fi(0, n) {
    int x = arr[i];
    ans = total - info[x][1] + (info[x][0] - 1) * (info[x][0] - 2) / 2;
    cout << ans << endl;
  }
  return 0;
}