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>
#define ll long long
#define ull unsigned long long
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(a,b) for(int a=0;a<(b);++a)
#define REP1(i,n) for(int i=1;i<=(n);++i)
#define debug(x) cerr<<#x<<": "<<x<<'\n'
#define all(x) (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define INF (1<<29)
#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
#define X first
#define Y second
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;


template<class T=int>
T in(){T x;cin>>x;return (x);}
template<class T>
void print(T& x){cout<<x<<'\n';}

const int MOD =(int)1e9+7;
const int mod =(int)1e9+7;
const int MAX =510000;
ll fac[MAX],finv[MAX],inv[MAX];
void COMint(){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}
ll COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0)return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}
ll gcd(ll a,ll b){
    if(b==0)return a;
    if(a>b){
        swap(a,b);
    }
    return gcd(a,b%a);
}
ll lcm(ll a,ll b){
    ll g;g=gcd(a,b);
    return a*b/g;
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}

bool compare_by_a(pair<int, int> a, pair<int, int> b) {
    if(a.first != b.first){
        return a.first < b.first;
    }else{
        return a.second < b.second;
    }
}
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 RS(ll N,ll P,ll M){
    if(P==0){
        return 1;
    }else{
        if(P%2==0){
            ll t=RS(N,P/2,M);
            return t*t%M;
        }else{
            return N*RS(N,P-1,M)%M;
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N=in();
    vector<int> a(N);
    map<int,int> mp;
    
    REP(i,N){
        cin>>a[i];
        mp[a[i]]+=1;
         
    }
    ll ans=0;
    for (auto iter=mp.begin();iter!=mp.end();iter++){
        int v=iter->second;
        ans+=v*(v-1)/2;
    }
    REP(i,N){
        if(mp[a[i]]>1){
            cout<<ans-(mp[a[i]]-1)<<endl;
        }else{
            cout<<ans<<endl;
        }
    }
    return 0;
}