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;

#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define endl '\n'
#define fcin ios::sync_with_stdio(false); cin.tie(nullptr);
#define INF 0x3f3f3f3f
#define all(x) x.begin(), x.end()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int, int> pii;

template <typename A, typename B>
string to_string(pair<A, B> p){
    return "(" + to_string(p.fi) + ", " + to_string(p.se) + ")";
}

template <typename T>
string to_string(T v){
    string s = "{";
    bool first = true;
    for(const auto & x : v){
        if(!first) s += ", ";
        first = false;
        s += to_string(x);
    }
    return s + "}";
}

void debug_out(){ cerr << endl;}
template <typename First, typename... Next>
void debug_out(First F, Next... N){
    cerr << " " << to_string(F);
    debug_out(N...);
}

#ifdef LOCAL
    #define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
    #define debug(...) 1
#endif

int main(){
    fcin;
    int n;
    cin >> n;
    vector<int> a(n);
    map<int, ll> f;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        f[a[i]]++;
    }
    ll ans = 0;
    for(auto p : f){
        ans += (p.se * (p.se - 1))/2;
    }

    for(int i = 0; i < n; i++){
        ll qt = f[a[i]];
        ll t = ans;
        t -= (qt * (qt-1))/2;
        qt--;
        t += max(0LL, (qt * (qt-1))/2);

        cout << t << endl;
    }
    return 0;
}