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)

// 単純に考えると
// 消す文字 N それぞれに対してヒストグラムを計算して ~ ってやって O(N**2)
// 間に合わなそうだからO(N) もしくは O(NlogN)で済ます方法を考える
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

const char wall = '#';
vector<vector<char>> fields;
vector<vector<bool>> used;
vector<vector<ll>> dist;
const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};

ll N;
vector<ll> A;

ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return b/gcd(a,b)*b;}

ll choose2(ll n){
  return n * (n-1)/2;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  // Globalなvectorのリサイズ
  //A.resize(N);
  
  cin >> N;
  
  vector<ll> a(N);
  rep(i, N) {
    cin >> a[i];
  }
  rep(i, N) {
    a[i]--;
  }
  vector<ll> cnt(N);
  rep(i, N) cnt[a[i]]++;
  
  ll tot = 0;
  // 何も削除しないとき答え
  rep(i, N) {
    tot += choose2(cnt[i]);
  }
  rep(i, N) {
    ll ans = tot;
    // i番目の数が減る
    // いったんa[i]をなかったことにして
    ans -= choose2(cnt[a[i]]);
    // 減った後の数で再計算
    ans += choose2(cnt[a[i]]-1);
    cout << ans << endl;
  }

  return 0;
}