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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384


int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;} 
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}    //最大公約数
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}     //最小公倍数
int nCr(int a,int b){
  if (b == 0 ){ return 0;}
  int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;} //コンビネーション
int nHr(int a,int b){return nCr(a+b-1,b);}       // 重複組み合わせ
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}  //階乗
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;} // a~bまでの階乗
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}   //数字の各位の和
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}  // b進数の各位の和？
int sankaku(int x){return ((1+x)*x)/2;} //三角数 xまでの和

//以下long long ver
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}

//double
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}


typedef pair<int, int> ii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
#define forr(i,a,b) ; for(int i=(a); i<(b); i++)
#define clean(arr,val) memset(arr,val,sizeof(arr))
#define forn(i,n) forr(i,0,n)
#define PB push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vpll;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
bool is_prime (int x){
  if (x<=1) return false ;
  for (int i=2 ; i*i<=x ; i++){
    if (x%i==0) return false ;
  }
   return true;
}
string alpha = "abcdefghijklmnopqrstuvwxyz" ;
string ALPHA = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" ;
    
bool chmin (int a,int b){
    if ( a > b ){
    a = b;
    return true ;
    }
    return false ;
}
 bool chmax (int a,int b){
    if ( a < b ){
    a = b;
    return true ;
    }
    return false ;
}   
#define SORT(c) sort((c).begin(),(c).end());
#define ANS(ans) cout<<(ans)<<endl;
 
 string split (string s , ll  n , char x){
   // n文字目は前の文字列
   string a ,b,c,d;
   a = s;
   b = s;
   if ( x == 'f'){
     return c  = a.erase(n) ;
   }else if ( x == 'l'){
     return d = b.substr(n) ;
   }
 }
/*CODE START HERE*/
int main() {
  int n;
  cin >> n;
  int memo[n+1];
  forn(i,n+1){
    memo[i] = 0;
  }
  vector<int> v(n)
    forn(i,n){
     cin >>  v.at(i) ;
      memo[v.at(i)] ++ ;
    }
    ll sum = 0;
    forn(i,n+1){
      if ( memo[i] >=2){
        sum += nCr(memo[i],2);
      }
    }
    forn(i,n+1){
      if (memo[v.at(i)] >=2){
        ANS(sum - memo[v.at(i)] -  1) ;
      }else{
        ANS(sum);
      }
    }
}